Problem set 33

1. The radius of the Fermi sphere of free electrons in a monovalent metal with an fcc structure, in which the volume of the unit cell is $a^3$, is
1. $\left(\frac{12\pi^2}{a^3}\right)^{1/3}$
2. $\left(\frac{3\pi^2}{a^3}\right)^{1/3}$
3. $\left(\frac{\pi^2}{a^3}\right)^{1/3}$
4. $\left(\frac{1}{a}\right)^{1/3}$
The radius of Fermi sphere is given by $k_f=\left(\frac{3\pi^2N}{V}\right)^{1/3}$. For FCC structure $\frac{N}{V}=\frac{4}{a^3}$. Hence, $k_f\left(\frac{12\pi^2}{a^3}\right)^{1/3}$. Hence, answer is (A).
2. Deviation from Rutherford scattering formula for $\alpha$-particle scattering gives an estimate of:
1. Size of an atom
2. Thickness of target
3. Size of a nucleus
4. half life of $\alpha$-emitter
Rutherford formula is derived by assuming nucleus to be point charge. However, nucleus is not a point charge. However, nucleus is not a point charge. There is a charge distribution inside the nucleus. Due to which there is change in angular dependence of scattering. From this information one can estimate size of nucleus.
Hence, answer is (C)
3. If the critical magnetic field for aluminium is $7.9\times 10^3 A/m$, the critical current which can flow through long thin superconducting wire of aluminium of diameter $1\times10^{-3}m$ is
1. 5.1 A
2. 1.6 A
3. 4.0 A
4. 2.48 A
Critical current is given by \begin{align*} I_c&=2\pi rH_c\\ &=2\times3.14*0.5\times10^{-3}\times7.9\times 10^3\\ &=2.48 A \end{align*} Hence, answer is (D)
4. An elemental dielectric has a dielectric constant $\epsilon=12$ and it contains $5\times10^{18} atoms/m^3$. Its electronic polarizability in uints of $FM^2$ is
1. $4.17\times10^{-30}$
2. $8.34\times10^{-30}$
3. $6.32\times10^{-30}$
4. $12.64\times10^{-30}$
The Clausius Mossotti relation is $$\frac{\epsilon_r-1}{\epsilon_r+2}=\frac{N\alpha_e}{3\epsilon_0}$$ \begin{align*} \alpha_e&=\frac{\epsilon_r-1}{\epsilon_r+2}\frac{3\epsilon_0}{N}\\ &=\frac{12-1}{12+2}\frac{3\times8.85\times10^{-12Fm^{-1}}}{5\times10^{18} atoms/m^3}\\ &=4.17\times10^{-30}FM^2\\ \end{align*} Hence, answer is (A)
5. Mangetite $(Fe_3O_4)$ has a cubic structure with a lattice constant of $8.4A^o$. The saturation magnetization in this material in units of A/m is :
1. $6.2\times10^5$
2. $6.2\times10^6$
3. $12.4\times10^6$
4. $12.4\times10^5$
As each $O$ has an oxidation number of $-2$ and there are 4 of them in the formula, the 3 Fe atoms must together have a charge of $+8$. This is consistent with their being two $Fe^{3+}$ ions and one $Fe^{2+}$ per formula unit. The ratio of $Fe^{3+}$ ions to $Fe^{2+}$ is 2 to 1. There are 8 $Fe_3O_4$ units per unit cell. Hence, there are 24 $Fe$ atoms in unit cell, out of which 16 are $Fe^{3+}$ and 8 are $Fe^{2+}$. Out of the 16 $Fe^{3+}$ ions 8 $Fe^{3+}$ ions are at octahedral position and 8 $Fe^{3+}$ ions are at tetrahedral position. The ions in tetrahedral sites oppose the applied magnetic field and the ions at octahedral positions reinforce the field. Consequently, $Fe^{3+}$ ions neutralizes the $Fe^{3+}$ at octahedral positions. In each $Fe^{3+}$ there are 5 unpaired $3d$ electrons and in each $Fe^{2+}$ there are 4 unpaired $3d$ electrons. Each unpaired electron has magnetic moment one Bohr magneton $\mu_B=9.27\times10^{-24}Am^2$. Hence, each $Fe^{2+}$ ion has 4$\mu_B$ magnetic moment. As, unit cell contains 8 $Fe^{2+}$ ions, total magnetic moment is $32\mu_B$.
Hence, saturation magnetization \begin{align*} M&=\frac{32\mu_B}{a^3}\\ &=\frac{32\times9.27\times10^{-24}}{\left(8.4\times10^{-10}\right)^3}\\ &=5.1\times10^5A/m \end{align*}

Problem set 32

1. The Hamiltonian of a particle of unit mass moving in the xy-plane is given to be: $H = xp_x- yp_y -\frac{1}{2}x^2 +\frac{1}{2} y^2$ in suitable units. The initial values are given to be $(x(0),y(0)) = (1,1)$ and $(p_x(0),p_y(0) )=(\frac{1}{2}, -\frac{1}{2})$. During the motion, the curves traced out by the particles in the xy-plane and the $p_xp_y$-plane are
1. both straight lines
2. a straight line and a hyperbola respectively
3. a hyperbola and an ellipse, respectively
4. both hyperbolas
$$\dot x=\frac{\partial H}{\partial p_x}=x$$ $$\dot p_x=-\frac{\partial H}{\partial x}=-(p_x-x)=x-p_x$$ $$\dot y=\frac{\partial H}{\partial p_y}=-y$$ $$\dot p_y=-\frac{\partial H}{\partial y}=-(-p_y+y)=p_y-y$$ Solving these equations and applying initial conditions, we get \begin{align*} x=e^x\\ y=e^{-x}\\ \end{align*} Or $$x=\frac{1}{y}$$ Which is equation of hyperbola. Similarly, solving equation for $p_x$ and $p_y$, one gets ellipse.
Hence, answer is (C)
2. A static, spherically symmetric charge distribution is given by $\rho(r) = \frac{A}{r} e^{-Kr}$ where $A$ and $K$ are positive constants. The electrostatic potential corresponding to this charge distribution varies with $r$ as
1. $re^{-Kr}$
2. $\frac{1}{r} e^{-Kr}$
3. $\frac{1}{r^2} e^{-Kr}$
4. $\frac{1}{r} (1-e^{-Kr})$
Since $\nabla^2V=-\rho/\epsilon_0$
$\nabla^2V$ must be proportional to $\frac{1}{r} e^{-Kr}$. Taking $\nabla^2=\frac{d^2}{dr^2}+\frac{2}{r}\frac{d}{dr}$, for the option (B) we get $\nabla^2\left(\frac{1}{r} e^{-Kr}\right)=\frac{K^2}{r} e^{-Kr}$
Hence, option (B) is correct.
3. Band-pass and band-reject filters can be implemented by combining a low pass and a high pass filter in series and in parallel, respectively. If the cut-off frequencies of the low pass and high pass filters are $\omega_0^{LP}$ and $\omega_0^{HP}$ , respectively, the condition required to implement the band-pass and band-reject filters are, respectively,
1. $\omega_0^{HP}<\omega_0^{LP}$ and $\omega_0^{HP}<\omega_0^{LP}$
2. $\omega_0^{HP}<\omega_0^{LP}$ and $\omega_0^{HP}>\omega_0^{LP}$
3. $\omega_0^{HP}>\omega_0^{LP}$ and $\omega_0^{HP}<\omega_0^{LP}$
4. $\omega_0^{HP}>\omega_0^{LP}$ and $\omega_0^{HP}>\omega_0^{LP}$
One can easily verify that the correct option is (B)
4. Non-interacting bosons undergo Bose-Einstein Condensation (BEC) when trapped in a three-dimensional isotropic simple harmonic potential. For BEC to occur, the chemical potential must be equal to
1. $\hbar\omega/2$
2. $\hbar\omega$
3. $3\hbar\omega/2$
4. $0$
For a three-dimensional isotropic simple harmonic potential, the lowest energy is $3\hbar\omega/2$ called as zero-point energy. Hence, option (C) is correct.
5. Linearly independent solution of the differential equation $$\frac{d^2y}{dx^2}+3\frac{dy}{dx}+2y=0$$ are:
1. $e^{-x}$, $e^{-2x}$
2. $e^{-x}$, $e^{2x}$
3. $e^{-2x}$, $e^{x}$
4. $e^{2x}$, $e^{x}$
Auxiliary equation is $$D^2+3D+2=0$$ $$(D+1)(D+2)=0$$ Hence, roots are $-1$ and $-2$

solutions are $e^{-x}$, $e^{-2x}$

Hence, answer is (A)

Problem set 31

1. The two dimensional lattice of graphene is an arrangement of Carbon atoms forming a honeycomb lattice of lattice spacing $a$, as shown below. The Carbon atoms occupy the vertices.
   
1. The Wigner-Seitz cell has an area of
1. $2a^2$
2. $\frac{\sqrt{3}}{2}a^2$
3. $6\sqrt{3}a^2$
4. $\frac{3\sqrt{3}}{2}a^2$
2. The Bravais lattice for this array is a
1. rectangular lattice with basis vectors $\vec d_1$ and $\vec d_2$
2. rectangular lattice with basis vectors $\vec c_1$ and $\vec c_2$
3. hexagonal lattice with basis vectors $\vec a_1$ and $\vec a_2$
4. hexagonal lattice with basis vectors $\vec b_1$ and $\vec b_2$
Wigner-Seitz cell contains one lattice point. Hence, volume of Wigner-Seitz cell is equal to volume of any normal cell divided by the number of lattice points in the normal cell ($N$). $${\scriptstyle\text{Area of Wigner-Seitz cell }=\frac{\text{area of normal cell}}{N}}$$ Here, normal cell is hexagonal, its area is $3\sqrt{3}a^2$ and number of lattice points $N=6\times\frac{1}{3}=2$. $$\text{Area of Wigner-Seitz cell }=\frac{3\sqrt{3}a^2}{2}$$
Hence, answer is (D)
The Bravais lattice for this array is a hexagonal lattice with basis vectors $\vec b_1$ and $\vec b_2$.
Hence, answer is (D)
2. A narrow beam of X-rays with wavelength $1.5 A^\circ$ is reflected from an ionic crystal with an fcc lattice structure with a density of $3.32 gm cm^{-3}$. The molecular weight is $108 AMU$ (1 AMU $= 1.66 \times10^{-24} g$).
1. The lattice constant is
1. $6.00A^\circ$
2. $4.56A^\circ$
3. $4.00A^\circ$
4. $2.56A^\circ$
2. The sine of the angle corresponding to $(111)$
1. $\sqrt{3}/4$
2. $\sqrt{3}/8$
3. $1/4$
4. $1/8$
For FCC structure number of lattice points per unit cell $n=4$ $$\rho=3.32 gm cm^{-3}=3320kg/m^3$$ $$M=108 AMU$$ Now, molecular weight is the weight of the 1 mole of molecules (1 mole of molecules means $N_0=6.02322\times10^{23}$ number of molecules i.e. Avogadro's number).
As, $N_0$ molecules have molecular weight $M$, $n$ molecules in the unit cell will have weight $\frac{n\times M}{N_0}$
$${\scriptstyle\text{Density of unit cell}=\frac{\text{weight of unit cell}}{\text{volume of unit cell}}}$$ $$\rho=\frac{n\times M}{N_0a^3}$$ \begin{align*} a^3&=\frac{n\times M}{N_0\rho}\\ &=\frac{4\times 108}{6.02322\times 10^{23}\times 3320}\\ &=6A^o \end{align*} Hence, answer is (A)
According to Bragg's law $$2d\sin\theta=\lambda\Rightarrow\sin\theta=\frac{\lambda}{2d}$$ $$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$$ For (111) plane $d=\frac{a}{\sqrt{3}}$ $$\sin\theta=\frac{\lambda\sqrt{3}}{2a}=\frac{1.5\times\sqrt{3}}{2\times6}=\frac{\sqrt{3}}{8}$$ Hence, answer is (B)
3. If an electron is in the ground state of the hydrogen atom, the probability that its distance from the proton is more than one Bohr radius is approximately
1. 0.68
2. 0.48
3. 0.28
4. 0.91
Ground state wave function is $\psi=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$. Probability that electron is found within Bohr radius is given by \begin{align*} &P(r)=\int\limits_0^{a_0}|\psi|^2\:4\pi r^2 dr\\ & =\frac{4}{a_0^3}\int\limits_0^{a_0}r^2e^{-2r/a_0}\:dr\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{\left[r^2\int e^{-2r/a_0}\:dr\right]_0^{a_0}-\left[\int 2r \left(\int e^{-2r/a_0}\:dr\right)\:dr\right]_0^{a_0}\right\}}\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{\left[r^2e^{-2r/a_0}\left(-\frac{a_0}{2}\right)\right]_0^{a_0}+a_0\left[\int r e^{-2r/a_0}\:dr\right]_0^{a_0}\right\}}\\ &{\scriptscriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}+a_0\left[r\int e^{-2r/a_0}\:dr-\int e^{-2r/a_0}\left(-\frac{a_0}{2}\right)\:dr\right]_0^{a_0}\right\}}\\ &{\scriptscriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}+a_0\left[re^{-2r/a_0}\left(-\frac{a_0}{2}\right)-e^{-2r/a_0}\left(\frac{a_0^2}{4}\right)\right]_0^{a_0}\right\}}\\ &{\scriptstyle=\frac{4}{a_0^3}\left\{-\frac{a_0^3e^{-2}}{2}-\frac{a_0^3e^{-2}}{2}-\frac{a_0^3e^{-2}}{4}+\frac{a_0^3}{4}\right\}}\\ &=-\left\{-5\times\frac{1}{e^2}+1\right\}=0.32 \end{align*} Hence, probability of finding electron outside Bohr radius=1-0.32=0.68
Hence, answer is (A)

Problem set 30

1. A particle is confined to the region $x \ge 0$ by a potential which increases linearly as $u(x) = u_0x$. The mean position of the particle at temperature $T$ is
1. $\frac{k_BT}{u_0}$
2. $\frac{(k_BT)^2}{u_0}$
3. $\sqrt{\frac{k_BT}{u_0}}$
4. $u_0k_BT$
$$ < x > = \frac{\int xe^{-\beta H}d\tau}{\int e^{-\beta H}d\tau}$$ where $\beta=\frac{1}{k_BT}$, and $H=\frac{p^2}{2m}+u(x)=\frac{p_x^2}{2m}+u_0x$ \begin{align*} &< x > =\frac{\int xe^{-\beta \left(\frac{p_x^2}{2m}+u_0x\right)}dxdp_x}{\int e^{-\beta \left(\frac{p_x^2}{2m}+u_0x\right)}dxdp_x}\\ &=\frac{\int_0^\infty xe^{-\beta u_0x}dx}{\int_0^\infty e^{-\beta u_0x}dx}\quad{\scriptstyle \int\limits_{0}^\infty x^{n}e^{-\alpha x}dx=\frac{n!}{\left(\alpha\right)^{n+1}}}\\ &=\frac{\frac{1}{(\beta u_0)^2}}{\frac{1}{\beta u_0}}={\frac{1}{\beta u_0}}={\frac{k_BT}{ u_0}} \end{align*} Hence, answer is (A).
2. A plane electromagnetic wave is propagating in a loss-less dielectric. The electric field is given by $${\scriptstyle\vec E(x,y,z,t)=E_0(\hat x+A\hat z)\exp{\left[ik_0\left\{-ct+\left(x+\sqrt{3}z\right)\right\}\right]}}$$ where $c$ is the speed of light in vacuum, $E_0$ , $A$ and $k_0$ are constants and $\hat x$ and $\hat z$ are unit vectors along the x- and z-, axes. The relative dielectric constant of the medium, $\epsilon_r$ and the constant $A$ are
1. $\epsilon_r=4$ and $A=-\frac{1}{\sqrt{3}}$
2. $\epsilon_r=4$ and $A=+\frac{1}{\sqrt{3}}$
3. $\epsilon_r=4$ and $A=\sqrt{3}$
4. $\epsilon_r=4$ and $A=-\sqrt{3}$
General form of wave equation is $\vec E(x,y,z,t)=E_0\hat ne^{i\left(\vec k\cdot\vec r-\omega t\right)}$. Comparing with this equation, we get $$\hat n=(\hat x+A\hat z)$$ \begin{align*} &\vec k\cdot\vec r=k_0\left(x+\sqrt{3}z\right)\\ &\Rightarrow\vec k=k_0\left(\hat x+\sqrt{3}\hat z\right) \end{align*} $$\omega=k_0c$$ $$v=\frac{\omega}{k}=\frac{k_0c}{k_0^2+3k_0^2}=\frac{c}{2}$$ $$\text{Refractive index } n=\frac{c}{v}=2$$ $$\text{Refractive index } n=\sqrt{\epsilon_2}\Rightarrow\epsilon_r=4$$ \begin{align*} &\text{Since }\vec k\cdot\hat n=0\\ &\Rightarrow k_0\left(\hat x+\sqrt{3}\hat z\right)\cdot(\hat x+A\hat z)\\ &=0 \end{align*} $$k_0\left(1+\sqrt{3}A\right)=0\Rightarrow A=-\frac{1}{\sqrt{3}}$$ Hence, answer is (A)
3. In a system consisting of two spin-$\frac{1}{2}$ particles labeled 1 and 2, let $\vec S^{(1)} = \frac{\hbar}{2}\vec\sigma^{(1)}$ and $\vec S^{(2)} = \frac{\hbar}{2}\vec\sigma^{(2)}$ denote the corresponding spin operators. Here $\vec\sigma\equiv (\sigma_x ,\sigma_y ,\sigma_z)$ and $\sigma_x ,\sigma_y ,\sigma_z$ are the three Pauli matrices.
1. In the standard basis the matrices for the operators $ S^{(1)}_x S^{(2)}_y$ and $S^{(1)}_y S^{(2)}_x$ are, respectively,
1. ${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad\frac{\hbar^2}{4}\begin{pmatrix}-1&0\\0&1\end{pmatrix}}$
2. ${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}i&0\\0&-i\end{pmatrix},\quad\frac{\hbar^2}{4}\begin{pmatrix}-i&0\\0&i\end{pmatrix}}$
3. $${\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}},$$$${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}}$$
4. $${\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&-i\\0&0&i&0\end{pmatrix}},$$$${\scriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0&-i&0&0\\i&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}}$$
2. These two operators satisfy the relation
1. ${\scriptstyle\left\{S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right\}=S^{(1)}_z S^{(2)}_z}$
2. ${\scriptstyle\left\{S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right\}=0}$
3. ${\scriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=iS^{(1)}_z S^{(2)}_z}$
4. ${\scriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=0}$
This is a double spin system, represented by $4\times4$ matrix representation. \begin{align*} S^{(1)}_x S^{(2)}_y=\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\otimes\frac{\hbar}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix} \end{align*} where, $\otimes$ represents tensor product. \begin{align*} &S^{(1)}_x S^{(2)}_y={\scriptscriptstyle\frac{\hbar^2}{4}\begin{pmatrix}0\begin{pmatrix}0&-i\\i&0\end{pmatrix}&1\begin{pmatrix}0&-i\\i&0\end{pmatrix}\\1\begin{pmatrix}0&-i\\i&0\end{pmatrix}&0\begin{pmatrix}0&-i\\i&0\end{pmatrix}\end{pmatrix}}\\ &= \frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix} \end{align*} \begin{align*} S^{(1)}_y S^{(2)}_x&=\frac{\hbar}{2}\begin{pmatrix}0&-i\\i&0\end{pmatrix} \otimes\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\\ &=\frac{\hbar^2}{4}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix} \end{align*} Hence, answer is (C) \begin{align*} &{\scriptscriptstyle\left[S^{(1)}_x S^{(2)}_y,S^{(1)}_y S^{(2)}_x\right]=S^{(1)}_x S^{(2)}_y\:S^{(1)}_y S^{(2)}_x-S^{(1)}_y S^{(2)}_x\:S^{(1)}_x S^{(2)}_y}\\ &{\scriptscriptstyle=\frac{\hbar^4}{16}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}}\\ &{\scriptscriptstyle-\frac{\hbar^4}{16}\begin{pmatrix}0&0&0&-i\\0&0&-i&0\\0&i&0&0\\i&0&0&0\end{pmatrix}\begin{pmatrix}0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0\end{pmatrix}}\\ &={\scriptscriptstyle\frac{\hbar^4}{16}\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{pmatrix}-\frac{\hbar^4}{16}\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{pmatrix}}\\ &=0 \end{align*} Hence, answer is (D)
4. The radius of $^{64}_{29}Cu$ nucleus is measured to be $4.8\times10^{-13} cm$.
1. The radius of $^{27}_{12}Mg$ nucleus can be estimated to be
1. $2.86\times10^{-13} cm$
2. $5.2\times10^{-13} cm$
3. $3.6\times10^{-13} cm$
4. $8.6\times10^{-13} cm$
2. The root-mean square (rms)- energy of a nucleon in a nucleus of atomic number $A$ in its ground state varies as:
1. $A^{4/3}$
2. $A^{1/3}$
3. $A^{-1/3}$
4. $A^{-2/3}$
Nuclear radius is given by $R=R_0A^{1/3}$ $$R_{Cu}=R_0A_{Cu}^{1/3}\quad \text{and}\quad R_{Mg}=R_0A_{Mg}^{1/3}$$ $$\frac{R_{Mg}}{R_{Cu}}=\left(\frac{A_{Mg}}{A_{Cu}}\right)^{1/3}=\left(\frac{27}{64}\right)^{1/3}=\frac{3}{4}$$ \begin{align*} R_{Mg}&=\frac{3}{4}R_{Cu}\\ &=\frac{3}{4}\times4.8\times10^{-13}\\ &=3.6\times10^{-13} cm \end{align*} Hence, answer is (C)
The root-mean square (rms)- energy of a nucleon in a nucleus of atomic number $A$ in its ground state varies as $A^{-1/3}$
Hence, answer is (C)

Problem set 29

1. Let $\psi_{nlm}$ denote the eigenstates of a hydrogen atom in the usual notation. The state $\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]$ is an eigenstate of
1. $L^2$ but not of the Hamiltonian or $L_z$
2. the Hamiltonian, but not of $L^2$ or $L_z$
3. the Hamiltonian, $L^2$ and $L_z$
4. $L^2$ and $L_z$, but not of the Hamiltonian
$${\scriptstyle\Psi=\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]}$$ As $\psi_{nlm}$ are eigenfunctions of $H$, their linear combination will be eigenfunction of $H$. Now, \begin{align*} L_z\psi_{nlm}&=m\hbar\psi_{nlm} \text{ and } L^2\psi_{nlm}\\ &=l(l+1)\hbar^2\psi_{nlm} \end{align*} \begin{align*} L_z\Psi&{\scriptstyle=L_z\frac{1}{5}\left[2\psi_{200}-3\psi_{211}+\sqrt{7}\psi_{210}-\sqrt{5}\psi_{21-1}\right]}\\ &=\frac{1}{5}\left[-3\psi_{211}+\sqrt{5}\psi_{21-1}\right] \end{align*} Hence, $\Psi$ is not eigenfunction of $L_z$. Similarly, one can show that $\Psi$ is not eigenfunction of $L^2$.
Hence, option (B) is correct.
2. The Hamiltonian for a spin-$1/2$ particle at rest is given by $H=E_0(\sigma_z+\alpha\sigma_x)$, where $\sigma_x$ and $\sigma_z$ are Pauli spin matrices and $E_0$ and $\alpha$ are constants. The eigenvalues of this Hamiltonian are
1. $\pm E_0\sqrt{1+\alpha^2}$
2. $\pm E_0\sqrt{1-\alpha^2}$
3. $E_0$ (doubly degenerate)
4. $E_0\left(1\pm\frac{1}{2}\alpha^2\right)$
\begin{align*} H&=E_0(\sigma_z+\alpha\sigma_x)\\ &=E_0\left\{\begin{pmatrix}1&0\\0&-1\end{pmatrix}+\alpha\begin{pmatrix}0&1\\1&0\end{pmatrix}\right\}\\ &=E_0\begin{pmatrix}1&\alpha\\\alpha&-1\end{pmatrix} \end{align*} Eigenvalues of $H$ are obtained by solving $\det|H-\lambda I|=0$ $$E_0\begin{vmatrix}1-\lambda&\alpha\\\alpha&-1-\lambda\end{vmatrix}=0$$ $$\lambda=\pm E_0\sqrt{1+\alpha^2}$$ Hence, answer is (A)
3. For a system of independent non-interacting one-dimensional oscillators, the value of the free energy per oscillator, in the limit $T\rightarrow0$, is
1. $\frac{1}{2}\hbar\omega$
2. $\hbar\omega$
3. $\frac{3}{2}\hbar\omega$
4. $0$
(A) $\frac{1}{2}\hbar\omega$
4. If the reverse bias voltage of a silicon varactor is increased by a factor of 2, the corresponding transition capacitance
1. increases by a factor of $\sqrt{2}$
2. increases by a factor of $2$
3. decreases by a factor of $\sqrt{2}$
4. decreases by a factor of $2$
\begin{align*} C_T&\propto\sqrt{\frac{1}{V}}\Rightarrow\frac{C_T'}{C_T}\\ &=\sqrt{\frac{V}{V'}}\\ &=\sqrt{\frac{V}{2V}}\\ &\Rightarrow C_T'=\frac{1}{\sqrt{2}}C_T \end{align*} Hence, answer is (C).
5. A cavity contains black body radiation in equilibrium at temperature $T$. The specific heat per unit volume of the photon gas in the cavity is of the form $C_v = \gamma T^3$,where $\gamma$ is a constant. The cavity is expanded to twice its original volume and then allowed to equilibrate at the same temperature $T$. The new internal energy per unit volume is
1. $4\gamma T^4$
2. $2\gamma T^4$
3. $\gamma T^4$
4. $\gamma T^4/4$
Change in internal energy $dU$ is given by $dU=C_vdT$ $$U=\int C_vdT=\int \gamma T^3dT=\frac{\gamma T^4}{4}$$ Hence, answer is (D)

Problem set 28

1. The solution of the differential equation $\frac{dx}{dt}=2\sqrt{1-x^2}$, with initial condition $x=0$ at $t=0$ is
1. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ \sinh{2t},\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
2. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{2}\\ 1,\quad t\geq\frac{\pi}{2} \end{cases} \end{align*}
3. \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{4}\\ 1,\quad t\geq\frac{\pi}{4} \end{cases} \end{align*}
4. $x=1-\cos{2t},\quad t\geq 0$
$$\int\frac{dx}{\sqrt{1-x^2}}=2\int dt$$ Let us make substitution $x=\sin\theta$, $dx=\cos\theta$ $$\int d\theta=2t+c_1$$ $$\theta=2t+c_1$$ $$\sin^{-1}x=c_1+2t$$ $$x=c+\sin{2t}$$ At $t=0$, $x=0$, hence, $c=0$ $$x=\sin{2t}$$ However, maximum value of $\sin(2t)$ is 1 $$\sin{2t}=1\quad \text{when } 2t=\frac{\pi}{2}\quad\text{ or }t=\frac{\pi}{4}$$ Hence, \begin{align*} x=\begin{cases} \sin{2t},\quad 0\leq t<\frac{\pi}{2}\\ 1,\quad t\geq\frac{\pi}{2} \end{cases} \end{align*} Hence, answer is (B).
2. Given a uniform magnetic field $\vec B=B_0\hat k$ (where $B_0$ is a constant), a possible choice for the magnetic vector potential $\vec A$ is
1. $B_0y\hat i$
2. $-B_0y\hat i$
3. $B_0(x\hat j+y\hat i)$
4. $B_0(x\hat i-y\hat j)$
Using, $\vec B=\vec\nabla \times\vec A$, for options
(A)$\vec B=\vec\nabla \times B_0y\hat i=-B_0\hat k$
(B)$\vec B=\vec\nabla \times (-B_0y\hat i)=B_0\hat k$
(C)$\vec B=\vec\nabla \times (B_0(x\hat j+y\hat i))=0$
(D)$\vec B=\vec\nabla \times (B_0(x\hat i-y\hat j))=2B_0\hat k$
Hence, answer is (B)
3. Consider a charge $Q$ at the origin of 3-dimensional coordinate system. The flux of the electric field through the curved surface of a cone that has a height $h$ and a circular base of radius $R$ is
1. $\frac{Q}{\epsilon_0}$
2. $\frac{Q}{2\epsilon_0}$
3. $\frac{hQ}{R\epsilon_0}$
4. $\frac{QR}{2h\epsilon_0}$
Let us consider a sphere of radius of $h$ (if $h>R$) or of radius $R$ (if $R>h$) with centre at origin. According to Gauss law the flux through entire sphere is $\frac{Q}{\epsilon_0}$. The flux through curved surfaces of cone is half of the flux through sphere i.e. $\frac{Q}{2\epsilon_0}$
Hence, answer is (B)
4. A Hermitian operator $\hat O$ has two normalised eigenstates $|1 > $ and $|2 > $ with eigenvalues $1$ and $2$, respectively. The two states $|u > =\cos\theta|1 > +\sin\theta|2 > $ and $|v > =\cos\phi|1 > +\sin\phi|2 > $ are such that $ < v|\hat O|v > =7/4$ and $ < u|v > =0$. Which of the following are possible values of $\theta$ and $\phi$?
1. $\theta=-\frac{\pi}{6}$ and $\phi=\frac{\pi}{3}$
2. $\theta=\frac{\pi}{6}$ and $\phi=\frac{\pi}{3}$
3. $\theta=-\frac{\pi}{4}$ and $\phi=\frac{\pi}{4}$
4. $\theta=\frac{\pi}{3}$ and $\phi=-\frac{\pi}{6}$
Let $\chi_1=|1 > $ and $\chi_2=|2 > $ \begin{align*} &< v|\hat O|v > =\int v*\hat Ovd\tau\\ &={\scriptstyle\int\left(\cos\phi\chi_1+\sin\phi\chi_2\right)^*\hat O\left(\cos\phi\chi_1+\sin\phi\chi_2\right)d\tau}\\ &={\scriptstyle\cos^2\phi\int\chi_1^*\hat O\chi_1 d\tau +\sin^2\phi\int\chi_2^*\hat O\chi_2d\tau}\\ &{\scriptstyle+\cos\phi\sin\phi\int\chi_1^*\hat O\chi_2d\tau+\cos\phi\sin\phi\int\chi_2^*\hat O\chi_1d\tau}\\ &=\cos^2\phi+2\sin^2\phi\\ &=1+\sin^2\phi \end{align*} $$1+\sin^2\phi=\frac{7}{4}$$ $$\Rightarrow \sin\phi=\frac{\sqrt{3}}{2}\Rightarrow\phi=\frac{\pi}{3}$$ \begin{align*} &< u|v > =\int u^*vd\tau\\ &{\scriptstyle=\int\left(\cos\theta\chi_1+\sin\theta\chi_2\right)^*\left(\cos\phi\chi_1+\sin\phi\chi_2\right)d\tau}\\ &={\scriptstyle\cos\theta\cos\phi\int\chi_1^*\chi_1 d\tau +\sin\theta\sin\phi\int\chi_2^*\chi_2d\tau}\\ &{\scriptstyle+\cos\theta\sin\phi\int\chi_1^*\chi_2d\tau+\sin\theta\sin\phi\int\chi_2^*\chi_1d\tau}\\ &=\cos\theta\cos\phi+\sin\theta\sin\phi\\ &=\cos{(\theta+\phi)}=\cos{(\theta+\frac{\pi}{3})} \end{align*} \begin{align*} &\Rightarrow \cos{(\theta+\frac{\pi}{3})}\\ &=0\Rightarrow\theta+\frac{\pi}{3}\\ &=\frac{\pi}{2}\Rightarrow\theta\\ =\frac{5\pi}{6}\text{ or }-\frac{\pi}{6} \end{align*} Hence, option (A) is correct.
5. The ground state energy of a particle of mass $m$ in the potential $V(x)=V_0\cosh{\left(\frac{x}{L}\right)}$, where $L$ and $V_0$ are constants (and $V_0>>\frac{\hbar^2}{2mL^2}$) is approximately
1. $V_0+\frac{\hbar}{L}\sqrt{\frac{2V_0}{m}}$
2. $V_0+\frac{\hbar}{L}\sqrt{\frac{V_0}{m}}$
3. $V_0+\frac{\hbar}{4L}\sqrt{\frac{V_0}{m}}$
4. $V_0+\frac{\hbar}{2L}\sqrt{\frac{V_0}{m}}$
\begin{align*} V(x)&=V_0\cosh{\left(\frac{x}{L}\right)}\\ &=\frac{V_0}{2}\left(e^{x/L}+ e^{-x/L}\right)\\ &=\frac{V_0}{2}\left\{\left[1+\frac{x}{L}+\frac{1}{2!}\left(\frac{x}{L}\right)^2+\cdots\right]\right.\\ &\left. +\left[1-\frac{x}{L}+\frac{1}{2!}\left(\frac{x}{L}\right)^2-\cdots\right]\right\}\\ &=\frac{V_0}{2}\left\{\left[2+2\frac{1}{2!}\left(\frac{x}{L}\right)^2+\cdots\right]\right\}\\ &=V_0+\frac{V_0}{2}\left(\frac{x}{L}\right)^2\\ &=V_0+\frac{1}{2}\frac{V_0}{L^2}x^2\\ &=V_0+\frac{1}{2}kx^2 \end{align*} where $k=\frac{V_0}{L^2}$. Thus, particle is under the action of constant potential $V_0$ and simple harmonic potential $\frac{1}{2}kx^2$. For free particle motion lowest energy is $V_0$ and for harmonic potential lowest energy is $\frac{1}{2}\hbar\omega$. Hence, ground state energy is \begin{align*} E&=V_0+\frac{1}{2}\hbar\omega\\ &=V_0+\frac{1}{2}\hbar\sqrt{\frac{k}{m}}\\ &=V_0+\frac{1}{2}\hbar\sqrt{\frac{V_0}{L^2m}} \end{align*} Hence, answer is (D)

Problem set 27

1. The Fourier transform of $f(x)$ is $\tilde{f}(k)=\int_{-\infty}^{\infty}dx\:e^{ikx}f(x)$. If $f(x)=\alpha\delta(x)+\beta\delta'(x)+\gamma\delta''(x)$, where $\delta(x)$ is the Dirac delta-function (and prime denotes derivative), what is $\tilde{f}(k)$?
1. $\alpha+i\beta k+i\gamma k^2$
2. $\alpha+\beta k-\gamma k^2$
3. $\alpha-i\beta k-\gamma k^2$
4. $i\alpha+\beta k-i\gamma k^2$
For Dirac delta function, we have, $$\int\limits f(x)\delta(x-y)\:dx=f(y)$$ $$\int\limits f(x)\delta'(x-y)\:dx=-f'(y)$$ $$\int\limits f(x)\delta''(x-y)\:dx=(-1)^2f''(y)$$ $$\int\limits f(x)\delta^n(x-y)\:dx=(-1)^nf^n(y)$$ \begin{align*} \tilde{f}(k)&={\scriptstyle\int\limits_{-\infty}^{\infty}dx\:e^{ikx}\left(\alpha\delta(x)+\beta\delta'(x)+\gamma\delta''(x)\right)}\\ &=\alpha f(0)+\beta(-f'(0))+\gamma f''(0)\\ &=\alpha-i\beta k-\gamma k^2 \end{align*} Hence, answer is (C)
2. A particle moves in three-dimensional space in a central potential $V(r)=kr^4$, where $k$ is a constant. The angular frequency $\omega$ for a circular orbit depends on its radius $R$ as
1. $\omega\propto R$
2. $\omega\propto R^{-1}$
3. $\omega\propto R^{1/4}$
4. $\omega\propto R^{-2/3}$
Force acting on particle is given by \begin{eqnarray} \begin{array}{c}\vec F(r)=-\nabla V(r)\\ {\scriptstyle=-\left(\hat r\frac{\partial}{\partial r}+\hat\theta\frac{1}{r}\frac{\partial}{\partial\theta}+\hat\phi\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\right)\left(kr^4\right)}\\ =-4kr^3\hat r\end{array} \end{eqnarray} For circular motion in radius $R$, we must have $$\frac{mv^2}{R}=F(R)=-4kR^3$$ Using $V=R\omega$ $$\frac{mR^2\omega^2}{R}=-4kR^3$$ $$\omega^2=\frac{-4kR^2}{m}$$ $$\omega\propto R$$ Hence, answer is (A)
3. The Lagrangian of a system is given by ${\scriptstyle L=\frac{1}{2}m\dot q_1^2+2m\dot q_2^2-k\left(\frac{5}{4}q_1^2+2q_2^2-2q_1q_2\right)}$ where $m$ and $k$ are positive constants. The frequencies of its normal modes are
1. $\sqrt{\frac{k}{2m}}\sqrt{\frac{3k}{m}}$
2. $\sqrt{\frac{k}{2m}}(13\pm\sqrt{73})$
3. $\sqrt{\frac{5k}{2m}}\sqrt{\frac{k}{m}}$
4. $\sqrt{\frac{k}{2m}}\sqrt{\frac{6k}{m}}$
Lagrange equation is $$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q_j}\right)-\frac{\partial L}{\partial q_j}$$ Substituting value of $L$ for generalized coordinates $q_1$ and $q_2$ we get $$m\ddot q_1+k\left(\frac{5}{2}q_1-2q_2\right)=0\quad (1)$$ $$4m\ddot q_2+k\left(4q_2-2q_1\right)=0\quad (2)$$ Form normal modes, both particles moves simple harmonically with same frequency $\omega$, hence, we have $$\ddot q_1=-\omega^2q_1$$ $$\ddot q_2=-\omega^2q_2$$ Substituting values of $\ddot q_1$ and $\ddot q_2$ in equations (1) and (2), we get $$\left(\frac{5}{2}k-m\omega^2\right)q_1=2kq_2\quad (3)$$ $$\left(4k-4m\omega^2\right)q_2=2kq_1\quad (4)$$ From equation (3) and (4) we get $$\frac{q_2}{q_1}=\frac{\left(\frac{5}{2}k-m\omega^2\right)}{2k}$$ $$\frac{q_2}{q_1}=\frac{2k}{\left(4k-4m\omega^2\right)}$$ Hence, we have $$\frac{\left(\frac{5}{2}k-m\omega^2\right)}{2k}=\frac{2k}{\left(4k-4m\omega^2\right)}$$ Solving this equation, we get $$2m^2\omega^4-7km\omega^2+3k^2=0$$ This is a quadratic equation in $\omega^2$. Solving this equation we get $$\omega^2=\frac{k}{2m}\quad\text{or }\frac{3k}{m}$$ or we get $$\omega=\sqrt{\frac{k}{2m}}\quad\text{or }\sqrt{\frac{3k}{m}}$$ Hence, answer is (A)
4. Consider a particle of mass $m$ moving with a speed $v$. If $T_R$ denotes the relativistic kinetic energy and $T_N$ its non-relativistic approximation, then the value of $(T_R-T_N)/T_R$ for $v=0.01\:c$, is
1. $1.25\times10^{-5}$
2. $5.0\times10^{-5}$
3. $7.5\times10^{-5}$
4. $1.0\times10^{-4}$
$$T_R=(\gamma-1)mc^2$$ where, $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=1.00005000375031$ $$T_R=5.000375031\times10^{-5}mc^2$$ $$T_N=0.5\times mv^2=5\times10^{-5}mc^2$$ $$\frac{(T_R-T_N)}{T_R}=7.5\times10^{-5}$$ Hence, answer is (C).
5. Two masses, $m$ each, are placed at the points $(x,y)=(a,a)$ and $(-a,-a)$. Two masses, $2m$ each, are placed at the points $(a,-a)$ and $(-a,a)$. The principal moments of inertia of the system are
1. $2ma^2$, $4ma^2$
2. $4ma^2$, $8ma^2$
3. $4ma^2$, $4ma^2$
4. $8ma^2$, $8ma^2$
Inertia tensor is given by $${\scriptscriptstyle{I}=\begin{pmatrix} \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2+x_{\alpha,3}^2\right)&-\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}&-\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,3}\\ -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}&\sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2+x_{\alpha,3}^2\right)&-\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,3}\\ -\sum\limits_\alpha m_\alpha x_{\alpha,3}x_{\alpha,1}&-\sum\limits_\alpha m_\alpha x_{\alpha,3}x_{\alpha,2}&\sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2+x_{\alpha,2}^2\right) \end{pmatrix} }$$ where, $\alpha$ runs over number of particles. Here, $x_{\alpha,1}$ means x-coordinate of particle number $\alpha$, $x_{\alpha,2}$ means y-coordinate of particle number $\alpha$ and $x_{\alpha,3}$ means z-coordinate of particle number $\alpha$

For a two-dimensional case, we have $${\scriptstyle{I}=\begin{pmatrix} \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2\right)&-\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\ -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}&\sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right) \end{pmatrix} }$$ \begin{align*} I_{11}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2\right)\\ &=ma^2+ma^2+2ma^2+2ma^2\\ &=6ma^2 \end{align*} \begin{align*} I_{12}&= -\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\ &=-(ma^2+ma^2-2ma^2-2ma^2)\\ &=2ma^2 \end{align*} \begin{align*} I_{21}&= -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}\\ &=-(ma^2+ma^2-2ma^2-2ma^2)\\ &=2ma^2 \end{align*} \begin{align*} I_{22}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right)\\ &=ma^2+ma^2+2ma^2+2ma^2\\ &=6ma^2 \end{align*} $${I}=\begin{pmatrix} 6ma^2&2ma^2\\ 2ma^2&6ma^2 \end{pmatrix} $$ The principal moments of inertia are obtained by solving $${I}=\begin{vmatrix} 6ma^2-\lambda&2ma^2\\ 2ma^2&6ma^2-\lambda \end{vmatrix}=0$$ Solving this determinant we get $$\lambda^2-12ma^2\lambda+32m^2a^4=0$$ Solving this quadratic equation in $\lambda$ we get, $\lambda_1=8ma^2$ and $\lambda_2=4ma^2$
Hence, answer is (B)

Problem set 26

1. The free energy difference between the superconducting and the normal states of a material is given by $\Delta F = F_s-F_N =\alpha |\psi|^2 + \frac{\beta}{2}|\psi|^4$, where $\psi$ is an order parameter and $\alpha$ and $\beta$ are constants such that $\alpha > 0$ in the normal and $\alpha < 0$ in the superconducting state, while $\beta > 0$ always. The minimum value of $\Delta F$ is
1. $-\alpha^2/\beta$
2. $-\alpha^2/2\beta$
3. $-3\alpha^2/\beta$
4. $-5\alpha^2/\beta$
For $\Delta F$ to be minimum $\frac{\partial(\Delta F)}{\partial\psi}=0$ i.e. \begin{align*} \frac{\partial(\Delta F)}{\partial\psi}&=2\alpha|\psi|+2\beta|\psi|^3\\ &=0\Rightarrow|\psi|^2\\ &=-\frac{\alpha}{\beta} \end{align*} \begin{align*} \Delta F&=-\alpha\times\frac{\alpha}{\beta}+\frac{\beta}{2}\frac{\alpha^2}{\beta^2}\\ &=-\frac{\alpha^2}{2\beta} \end{align*} Hence, answer is (B)
2. Consider a hydrogen atom undergoing a $2P\rightarrow 1S$ transition. The lifetime $t_{sp}$ of the $2P$ state for spontaneous emission is $1.6 ns$ and the energy difference between the levels is $10.2eV$. Assuming that the refractive index of the medium $n_0 = 1$, the ratio of Einstein coefficients for stimulated and spontaneous emission $B_{21}(\omega)/A_{21}(\omega)$ is given by
1. $0.683\times10^{12} m^3J^{-1}s^{-1}$
2. $0.146\times10^{-12} Jsm^{-3}$
3. $6.83\times10^{12} m^3J^{-1}s^{-1}$
4. $1.46\times10^{-12} Jsm^{-3}$
\begin{align*} \frac{B_{21}(\omega)}{A_{21}(\omega)}&=\frac{\pi^2c^3}{\hbar\omega^3n_0^3}\\ &=\frac{\pi^2c^3\hbar^2}{(\Delta E)^3n_0^3}\\ &=\frac{c^3h^2}{4(\Delta E)^3n_0^3} \end{align*} $\Delta E=10.2 eV$, $n_0=1$ \begin{eqnarray} \begin{array}{c}\frac{B_{21}(\omega)}{A_{21}(\omega)}=\frac{\left(3\times10^8\right)^3\times\left(6.63\times10^{-34}\right)^2}{\left(10.2\times1.6\times10^{-19}\right)^3\times1}\\ =0.683\times10^{12}m^3J^{-1}s^{-1}\end{array} \end{eqnarray} Hence, answer is (A)
3. In the scattering of some elementary particles, the scattering cross-section is found to depend on the total energy and the fundamental constants $h$ (Planck’s constant) and $c$ (the speed of light in vacuum). Using dimensional analysis, the dependence of $\sigma$ on these quantities is given by
1. $\sqrt{\frac{hc}{E}}$
2. $\frac{hc}{E^{3/2}}$
3. $\left(\frac{hc}{E}\right)^2$
4. $\frac{hc}{E}$
Scattering cross-section is measured in units of area i.e. $m^2$ $$\frac{hc}{E}=\frac{Js\:m/s}{J}=m$$ $$\left(\frac{hc}{E}\right)^2=m^2$$ Hence, answer is (C)
4. If $y=\frac{1}{\tanh x}$, then $x$ is
1. $\ln{\left(\frac{y+1}{y-1}\right)}$
2. $\ln{\left(\frac{y-1}{y+1}\right)}$
3. $\ln{\sqrt{\frac{y-1}{y+1}}}$
4. $\ln{\sqrt{\frac{y+1}{y-1}}}$
$$\sinh{x}=\frac{e^x-e^{-x}}{2}$$ $$\cosh{x}=\frac{e^x+e^{-x}}{2}$$ $$\tanh{x}=\frac{\sinh{x}}{\cosh{x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ $$y=\frac{1}{\tanh{x}}=\frac{e^x+e^{-x}}{e^x-e^{-x}}$$ $$\frac{y+1}{y-1}=\frac{\frac{e^x+e^{-x}}{e^x-e^{-x}}+1}{\frac{e^x+e^{-x}}{e^x-e^{-x}}-1}=e^{2x}$$ $$2x=\ln{\left(\frac{y+1}{y-1}\right)}$$ $$x=\ln{\sqrt{\frac{y+1}{y-1}}}$$ Hence, answer is (D)
5. The function $\frac{z}{\sin{\pi z^2}}$ of a complex variable $z$ has
1. a simple pole at $0$ and poles of order $2$ at $\pm\sqrt{n}$ for $n=1,2,3,\dots$
2. a simple pole at $0$ and poles of order $2$ at $\pm\sqrt{n}$ and $\pm i\sqrt{n}$ for $n=1,2,3,\dots$
3. poles of order $2$ at $\pm\sqrt{n}$ for $n=0,1,2,3,\dots$
4. poles of order $2$ at $\pm n$ for $n=0,1,2,3,\dots$
$$\sin{\pm n\pi}=0,\quad \text{for } n=0,1,2,\dots$$ $$z^2=\pm n$$ $$z=\pm\sqrt{n} \quad\text{or } z=\pm i\sqrt{n}$$ Hence, answer is (B)

Problem set 25

1. The energies in the ground state and first excited state of a particle of mass $m =\frac{1}{2}$ in a potential $V(x)$ are $-4$ and $-1$, respectively, (in units in which $\hbar = 1$ ). If the corresponding wavefunctions are related by $\psi_1(x)=\psi_0(x) \sinh x$, then the ground state eigenfunction is
1. $\psi_0(x)=\sqrt{sech~ x}$
2. $\psi_0(x)=sech~ x$
3. $\psi_0(x)=sech^2~ x$
4. $\psi_0(x)=sech^3~ x$
Schrodinger equation is $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$$ For $\hbar=1$,$m=m/2$ $$-\frac{d^2\psi}{dx^2}+V\psi=E\psi$$ For ground state $$-\frac{d^2\psi_0}{dx^2}+V\psi_0=-4\psi_0\quad \dots(1)$$ $$-\frac{d^2\psi_1}{dx^2}+V\psi_1=-\psi_1\quad \dots(2)$$ Substituting $\psi_1=\psi_0\sinh{x}$ in equation (2) $${\scriptstyle-\frac{d^2(\psi_0\sinh{x})}{dx^2}+V\psi_0\sinh{x}=-\psi_0\sinh{x}}$$ Solving this we get $$ {\scriptstyle-\frac{d^2\psi_0}{dx^2}\sinh{x}-2\frac{d\psi_0}{dx}\cosh{x}-\psi_0\sinh{x}+V\psi_0=-\psi_0\sinh{x}} $$ Rearranging \begin{align*} &-\frac{d^2\psi_0}{dx^2}-2\coth{x}\frac{d\psi_0}{dx}-\psi_0\sinh{x}\\ &+V\psi_0=-\psi_0\sinh{x} \end{align*} substituting value of $-\frac{d^2\psi_0}{dx^2}+V\psi_0=-4\psi_0$ from equation (1), we get $$\coth{x}\frac{d\psi_0}{dx}=-2\psi_0$$ $$\frac{d\psi_0}{\psi_0}=-2\frac{dx}{\coth{x}}=-2\tanh{x}dx$$ Integrating above equation $$\ln\psi_0=-2\ln{\cosh{x}}=\ln\text{sech}^{2}{x}$$ $$\psi_0=\text{sech}^{2}{x}$$ Hence, answer is (C)
2. The perturbation \begin{align*} H'=\begin{cases} b(a-x),&-a < x < a\\ 0&\text{otherwise} \end{cases} \end{align*} acts on a particle of mass $m$ confined in an infinite square well potential \begin{align*} V(x)=\begin{cases} 0,&-a < x < a\\ \infty&\text{otherwise} \end{cases} \end{align*} The first order correction to the ground-state energy of the particle is
1. $\frac{ba}{2}$
2. $\frac{ba}{\sqrt{2}}$
3. $2ba$
4. $ba$
\begin{align*} &E_0^{(1)}=\int\psi^*H'\psi\:dx\\ &=\frac{1}{a}\int\limits_{-a}^{a}\sin{\frac{\pi x}{2a}}b(a-x)\sin{\frac{\pi x}{2a}}dx\\ &=\frac{b}{a}\int\limits_{-a}^{a}\sin^2{\frac{\pi x}{2a}}\:(a-x)\:dx\\ &=\frac{b}{2a}\int\limits_{-a}^{a}\left(1-\cos{\frac{\pi x}{a}}\right)(a-x)dx\\ &=\frac{b}{2a}\left\{\int\limits_{-a}^{a}(a-x)dx\right.\\ &\left.-\int\limits_{-a}^{a}\cos{\frac{\pi x}{a}}(a-x)dx\right\} \end{align*} Odd functions have integration zero \begin{align*} E_0^{(1)}&=\frac{b}{2a}\left\{a\int\limits_{-a}^{a}dx-a\int\limits_{-a}^{a}\cos{\frac{\pi x}{a}}dx\right\}\\ &=\frac{b}{2a}\left\{2a^2-0\right\}=ba \end{align*} Hence, answer is (D)
3. Let $|0>$ and $|1>$ denote the normalized eigenstates corresponding to the ground and the first excited states of a one-dimensional harmonic oscillator. The uncertainty $\Delta x$ in the state $\frac{1}{\sqrt{2}}\left(|0>+|1>\right)$ is
1. $\Delta x=\sqrt{\hbar/2m\omega}$
2. $\Delta x=\sqrt{\hbar/m\omega}$
3. $\Delta x=\sqrt{2\hbar/m\omega}$
4. $\Delta x=\sqrt{\hbar/4m\omega}$
$$\Delta x=\sqrt{ < x^2 > - < x >^2}$$ Let $u_0=|0 > $, $u_1=|1 > $ and $\psi= |0 > +|1 > =u_0+u_1$. The lowering and raising operators for harmonic oscillator are $$\hat a=\frac{1}{\sqrt{2m\hbar\omega}}\left(m\omega x+ip\right)$$ $$\hat a^\dagger=\frac{1}{\sqrt{2m\hbar\omega}}\left(m\omega x-ip\right)$$ $$x=\sqrt{\frac{\hbar}{2m\omega}}\left(\hat a+\hat a^\dagger\right)$$ Also we have, $\hat a u_n=\sqrt{n}u_{n-1}$, $\hat a^\dagger u_n=\sqrt{n+1}u_{n+1}$. \begin{align*} &< x >=<\psi|x|\psi >\\ &={\scriptstyle\sqrt{\frac{\hbar}{2m\omega}}<\psi|\left(\hat a+\hat a^\dagger\right)|\psi >}\\ &={\scriptstyle\sqrt{\frac{\hbar}{2m\omega}}\frac{1}{2} < (u_0+u_1)|\left(\hat a+\hat a^\dagger\right)|(u_0+u_1) >}\\ &={\scriptstyle\sqrt{\frac{\hbar}{8m\omega}}\left\{ < u_0|\hat a|u_0 > + < u_0|\hat a| u_1 >+ < u_1|\hat a|u_0>\right.}\\ &{\scriptstyle+ < u_1|\hat a|u_1 >+ < u_0|\hat a^\dagger|u_0 > + < u_0|\hat a^\dagger|u_1 >}\\ &{\scriptstyle\left.+ < u_1|\hat a^\dagger|u_0 > + < u_1|\hat a^\dagger|u_1 >\right\}}\\ &={\scriptstyle\sqrt{\frac{\hbar}{8m\omega}}\left\{0+1+0+0+0+0+1+0\right\}}\\ &=\sqrt{\frac{\hbar}{2m\omega}} \end{align*} \begin{align*} &< x^2 > ={\scriptstyle < \psi|x^2|\psi > =\frac{\hbar}{2m\omega} < \psi|\left(\hat a+\hat a^\dagger\right)^2|\psi >} \\ &={\scriptstyle\frac{\hbar}{2m\omega}\frac{1}{2}<(u_0+u_1)|\left(\hat a+\hat a^\dagger\right)^2|(u_0+u_1) > }\\ &={\scriptstyle\frac{\hbar}{4m\omega} < (u_0+u_1)|\left(\hat a\hat a+\hat a^\dagger\hat a^\dagger+\hat a\hat a^\dagger+\hat a^\dagger\hat a\right)|(u_0+u_1) > }\\ &={\scriptstyle\frac{\hbar}{4m\omega}\left\{ < u_0|\hat a\hat a|u_0 > + < u_0|\hat a\hat a|u_1 > + < u_1|\hat a\hat a|u_0 >\right.}\\ &{\scriptstyle + < u_1|\hat a\hat a|u_1 > + < u_0|\hat a^\dagger\hat a^\dagger|u_0 > + < u_0|\hat a^\dagger\hat a^\dagger|u_1 >}\\ &{\scriptstyle + < u_1|\hat a^\dagger\hat a^\dagger|u_0>+ < u_1|\hat a^\dagger\hat a^\dagger|u_1 >+ < u_0|\hat a\hat a^\dagger+\hat a^\dagger\hat a|u_0 >}\\ &{\scriptstyle + < u_0|\hat a\hat a^\dagger+\hat a^\dagger\hat a|u_1 > + < u_1|\hat a\hat a^\dagger+\hat a^\dagger\hat a|u_0 >}\\ &{\scriptstyle\left. + < u_1|\hat a\hat a^\dagger+\hat a^\dagger\hat a|u_1 > \right\}} \end{align*} Using identities $\hat a\hat a^\dagger+\hat a^\dagger\hat a=(2n+1)u_n$ and mentioned above, we get $$ < x^2 > =\frac{\hbar}{m\omega}$$ $$\Delta x=\sqrt{\frac{\hbar}{m\omega}-\frac{\hbar}{2m\omega}}=\sqrt{\frac{\hbar}{2m\omega}}$$
4. What would be the ground state energy of the Hamiltonian $H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}-\alpha \delta(x)$ if variational principle is used to estimate it with the trial wavefunction $\psi(x)= Ae^{-bx^2}$ with $b$ as the variational parameter? [Hint: ${\scriptstyle\int_{-\infty}^{\infty} x^{2n} e^{-2bx^2}\:dx = (2b)^{-n-\frac{1}{2}}\Gamma\left(n+\frac{1}{2}\right)}$]
1. $-m\alpha^2/2\hbar^2$
2. $-2m\alpha^2/\pi\hbar^2$
3. $-m\alpha^2/\pi\hbar^2$
4. $m\alpha^2/\pi\hbar^2$
According to variational principle ground state energy is given by $$E=\frac{<\psi|H|\psi>}{<\psi|\psi>}$$ \begin{align*} <\psi|\psi>&=|A|^2\int\limits_{-\infty}^{\infty}e^{-2bx^2}\:dx\\ &=|A|^2\sqrt{\frac{\pi}{2b}} \end{align*} Using normalization condition $<\psi|\psi>=1$, we get, $A=\left(\frac{2b}{\pi}\right)^{\frac{1}{4}}$ $$\psi(x)= \left(\frac{2b}{\pi}\right)^{\frac{1}{4}}e^{-bx^2}$$ \begin{align*} & E=<\psi|H|\psi>\\ &=\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:H\:e^{-bx^2}\:dx\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}-\alpha \delta(x)\right)\:e^{-bx^2}\:dx}\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{-\frac{\hbar^2}{2m}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:\frac{d^2}{dx^2}e^{-bx^2}dx\right.}\\ &{\scriptstyle\left.-\int\limits_{-\infty}^{\infty}e^{-bx^2}\alpha \delta(x)\:e^{-bx^2}\:dx\right\}}\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{-\frac{\hbar^2}{2m}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:\frac{d}{dx}\left(-2bxe^{-bx^2}\right)dx\right.}\\ &{\scriptstyle\left.-\alpha\int\limits_{-\infty}^{\infty}e^{-2bx^2} \delta(x)\:dx\right\}}\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{\frac{\hbar^2b}{m}\int\limits_{-\infty}^{\infty}e^{-bx^2}\:\left(e^{-bx^2}-2bxe^{-bx^2}\right)dx-\alpha\right\}}\\ &={\scriptstyle\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{\frac{\hbar^2b}{m}\left[\int\limits_{-\infty}^{\infty}e^{-2bx^2}\:dx-2b\int\limits_{-\infty}^{\infty}xe^{-2bx^2}dx\right]-\alpha\right\}}\\ &=\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\left\{\frac{\hbar^2b}{m}\left[\sqrt{\frac{\pi}{2b}}-0\right]-\alpha\right\}\\ &=\frac{\hbar^2b}{m}-\left(\frac{2b}{\pi}\right)^{\frac{1}{2}}\alpha \end{align*} $$\frac{\partial E}{\partial b}=\frac{\hbar^2}{m}-\frac{1}{2}\left(\frac{2}{b\pi}\right)^{\frac{1}{2}}\alpha=0$$ gives $$b=\frac{m^2\alpha^2}{2\hbar^4\pi}$$ \begin{align*} E&=\frac{\hbar^2}{m}\frac{m^2\alpha^2}{2\hbar^4\pi}-\left(\frac{2}{\pi}\frac{m^2\alpha^2}{2\hbar^4\pi}\right)^{\frac{1}{2}}\alpha\\ &=-\frac{m\alpha^2}{2\pi\hbar^2} \end{align*}
5. A given quantity of gas is taken from the state $A \rightarrow C$ reversibly, by two paths, $A\rightarrow C$ directly and $A\rightarrow B\rightarrow C$ as shown in the figure below.
   
   
   
   During the process $A\rightarrow C$ the work done by the gas is $100 J$ and the heat absorbed is $150 J$. If during the process $A\rightarrow B\rightarrow C$ the workdone by the gas is $30 J$, the heat absorbed is
1. 20 J
2. 80 J
3. 220 J
4. 280 J
$$\delta Q=dU+\delta W$$ For path $A\rightarrow C$, $\delta W=100 J$ and $\delta Q=150 J$ $$150=du+100$$ $$dU=50 J$$ As, $dU$ is a state function, it does not depend on the path followed. Hence, for the path $A\rightarrow B\rightarrow C$, we have $$\delta Q=50+30=80 J$$ Hence, answer is (B)

Problem set 24

1. The Hamiltonian of a simple pendulum consisting of a mass $m$ attached to a massless string of length $l$ is $H = \frac{p_\theta^2}{2ml^2}+ mgl(1- \cos\theta)$. If $L$ denotes the Lagrangian, the value $\frac{dL}{dt}$ is:
1. $-\frac{2g}{l}p_\theta\sin\theta$
2. $-\frac{g}{l}p_\theta\sin{2\theta}$
3. $\frac{g}{l}p_\theta\cos{\theta}$
4. $lp_\theta^2\cos{\theta}$
$$L=\sum\limits_ip_iq_i-H$$ $$L=p_\theta\dot\theta-\frac{p_\theta^2}{2ml^2}- mgl(1- \cos\theta)$$ As $L\equiv L(q,\dot q,t)$ $$\frac{dL}{dt}=\frac{\partial L}{\partial q}\frac{\partial q}{\partial t}+\frac{\partial L}{\partial\dot q}\frac{\partial\dot q}{\partial t}+\frac{\partial L}{\partial t}$$ In our case$L\equiv L(\theta,\dot \theta)$ $$\frac{dL}{dt}=\frac{\partial L}{\partial \theta}\frac{\partial \theta}{\partial t}+\frac{\partial L}{\partial\dot\theta}\frac{\partial\dot\theta}{\partial t}$$ $$\frac{\partial L}{\partial \theta}=-mgl\sin{\theta}$$ $$\frac{\partial L}{\partial\dot \theta}=p_\theta$$ $$\frac{dL}{dt}=-mgl\sin{\theta}\: \dot\theta+p_\theta\:\ddot\theta$$ $$\dot\theta=\frac{\partial H}{\partial p_\theta}=\frac{p_\theta}{ml^2}$$ $$\ddot\theta=\frac{\dot p_\theta}{ml^2}$$ But $$\dot p_\theta=-\frac{\partial H}{\partial \theta}=-mgl\sin{\theta}$$ $$\ddot\theta=-\frac{g}{l}\sin{\theta}$$ $$\frac{dL}{dt}=-\frac{2g}{l}p_\theta\sin{\theta}$$ Hence, answer is (A)
2. Two bodies of equal mass $m$ are connected by a massless rigid rod of length $l$ lying in the XY-plane with the centre of the rod at the origin. If this system is rotating about the $z$-axis with a frequency $\omega$, its angular momentum is
1. $ml^2\omega/4$
2. $ml^2\omega/2$
3. $ml^2\omega$
4. $2ml^2\omega$
Angular momentum, $L=I\omega$ $$I=\sum_im_ir_i^2$$ Here, $r_i=\frac{l}{2}$ $$I=m\frac{l^2}{4}+m\frac{l^2}{4}=m\frac{l^2}{2}$$ $$L=\frac{ml^2\omega}{2}$$ Hence, answer is (B)
3. An infinite solenoid with its axis of symmetry along the z-direction carries a steady current $I$. The vector potential $\vec A$ at a distance $R$ from the axis
1. is constant inside and varies as $R$ outside the solenoid
2. varies as $R$ inside and is constant outside the solenoid
3. varies as $\frac{1}{R}$ inside and as $R$ outside the solenoid
4. varies as $R$ inside and as $\frac{1}{R}$ outside the solenoid
Let $r$ is the radius of solenoid.
Magnetic filed inside the solenoid is $B=\mu_0 n I$, where $n$ is number of turns per unit length, and, $I$ is current flowing through the wire. Magnetic filed outside the solenoid is zero.

Vector potential inside the solenoid:
Now, according to Stoke's law, we have $$\oint \vec A\cdot d\vec l=\int_S\left(\nabla\times\vec A\right)\cdot d\vec S$$ But $\left(\nabla\times\vec A\right)=\vec B$ $$\oint \vec A\cdot d\vec l=\int_S\vec B\cdot d\vec S=\Phi_B=flux$$ Let us calculate flux $\Phi_B$. For this let us consider a circular path of radius $R$ centered along its axis. The flux through the circular area is $$\Phi_B=\pi R^2B=\pi R^2\mu_0 n I$$ Hence, $$\oint \vec A\cdot d\vec l=\pi R^2\mu_0 n I$$ But, $\oint \vec A\cdot d\vec l=2\pi R A_\Phi$, this is because $\vec A$ is along the direction of current which is circumferential. $$2\pi R A_\Phi=\pi R^2\mu_0 n I$$ $$A_\Phi=\mu_0 n I\frac{R}{2}$$

Vector potential outside the solenoid:
Let us consider the point $R>r$. The flux through circular area is $$\Phi_B=\pi r^2B=\pi r^2\mu_0 n I$$ $$\oint \vec A\cdot d\vec l=\pi r^2\mu_0 n I$$ But, $\oint \vec A\cdot d\vec l=2\pi R A_\Phi$ $$2\pi R A_\Phi=\pi r^2\mu_0 n I$$ $$A=\mu_0 n I\frac{r^2}{2R}$$
Hence, $\vec A$ varies as $R$ inside and as $\frac{1}{R}$ outside the solenoid
Hence, answer is (D)
4. Consider an infinite conducting sheet in the xy-plane with a time dependent cunent density $Kt\hat i$, where $K$ is a constant. The vector potential at $(x, y,z)$ is given by $\vec A=\frac{\mu_0K}{4c}(ct-z)^2\hat i$. The magnetic field $\vec B$ is
1. $\frac{\mu_0Kt}{2}\hat j$
2. $-\frac{\mu_0Kz}{2c}\hat j$
3. $-\frac{\mu_0K}{2c}(ct-z)\hat i$
4. $-\frac{\mu_0K}{2c}(ct-z)\hat j$
$$\vec A=\frac{\mu_0K}{4c}(ct-z)^2\hat i$$ \begin{align*} \vec B&=\vec\nabla\times\vec A\\ &=\begin{vmatrix} \hat i&\hat j&\hat k\\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \frac{\mu_0K}{4c}(ct-z)^2&0&0 \end{vmatrix} \end{align*} $$\vec B=-\frac{\mu_0K}{2c}(ct-z)\hat j$$ Hence, answer is (D)
5. When a charged particle emits electromagnetic radiation, the electric field $\vec E$ and the Poynting vector $\vec S = \frac{1}{\mu_0} \vec E\times\vec B$ at a large distance $r$ from the emitter vary as $\frac{1}{r^n}$ and $\frac{1}{r^m}$ respectively. Which of the following choices for $n$ and $m$ are correct?
1. $n = 1$ and $m=1$
2. $n = 2$ and $m=2$
3. $n = 1$ and $m=2$
4. $n = 2$ and $m=4$
When a charged particle emits electromagnetic radiation $E\propto\frac{1}{r}$ and $B\propto\frac{1}{r}$. As $\vec S=\frac{\vec E\times \vec B}{\mu_0}$, $S\propto\frac{1}{r^2}$
Hence, answer is (C)

Problem set 23

1. A diode $D$ as shown in the circuit has an $i-V$ relation that can be approximated by \begin{align*} i_{_D}=\begin{cases} v^2_{_D}+2v_{_D},&\text{for }v_{_D}>0\\ 0,&\text{for }v_{_D}\leq 0 \end{cases} \end{align*}
   
   The value of $v_{_D}$ in the circuit is
1. $(-1+\sqrt{11})$
2. $8V$
3. $5V$
4. $2V$
Applying Kirchhoff's voltage law $$i_{_D}(1)+v_{_D}=10$$ Also, for $v_{_D}>0$ we have $$i_{_D}=v^2_{_D}+2v_{_D}$$ Substituting value of $i_{_D}$ from second equation into the first equation, we get $$v^2_{_D}+3v_{_D}-10=0$$ Solving this quadratic equation for $v_{_D}$, we get, $v_{_D}=2$ or $v_{_D}=-5$. But for $v_{_D}<0$ we get $i_{_D}=0$, hence $v_{_D}=2$.
Hence, answer is (D)
2. The Taylor expansion of the function $\ln{(\cosh x)}$, where $x$ is real, about the point $x= 0$ starts with the following terms:
1. $-\frac{1}{2}x^2+\frac{1}{12}x^4+\cdots$
2. $\frac{1}{2}x^2-\frac{1}{12}x^4+\cdots$
3. $-\frac{1}{2}x^2+\frac{1}{6}x^4+\cdots$
4. $\frac{1}{2}x^2+\frac{1}{6}x^4+\cdots$
We will require following identities: $$\sinh{x}=\frac{e^x-e^{-x}}{2}$$ $$\cosh{x}=\frac{e^x+e^{-x}}{2}$$ $$\tanh{x}=\frac{\sinh{x}}{\cosh{x}}$$ $$\frac{d}{dx}\sinh{x}=\cosh{x}$$ $$\frac{d}{dx}\cosh{x}=\sinh{x}$$ $$\frac{d}{dx}\tanh{x}=\text{sech }^2{x}$$ $$\frac{d}{dx}\text{sech }{x}=-\text{sech }{x}\tanh{x}$$ Taylor series of $f(x)$ about $x=0$ is given by \begin{align*} f(x)&=f(x)+\frac{1}{1!}f'(0)x+\frac{1}{2!}f''(0)x^2\\ &+\frac{1}{3!}f'''(0)x^3+\frac{1}{4!}f''''(0)x^4+.. \end{align*} \begin{align*} &f(x)=\ln{(\cosh{x})}\\ &\Rightarrow f(0)=0\\ &f'(x)=\tanh{x} \\ &\Rightarrow f'(0)=0\\ &f''(x)=\text{sech }^2{x}=1-\tanh^2{x}\\ &\Rightarrow f''(0)=1\\ &f'''(x)=-2\text{sech }^2{x}\tanh{x}\\ &\Rightarrow f'''(0)=0\\ f''''(x)&=4\text{sech }^2{x}\tanh^2{x}-2\text{sech }^4{x} \\ &\Rightarrow f''''(0)=-2 \end{align*} $$f(x)=\frac{1}{2!}x^2-\frac{1}{12}x^4+\cdots$$ Hence, answer is (B)
3. Given $2\times2$ unitary matrix $U$ satisfying $U^\dagger U=UU^\dagger = 1$ with $\det U = e^{i\phi}$, one can construct a unitary matrix $V \:( V^\dagger V = VV^\dagger= 1)$ with $\det V=1$ from it by
1. multiplying $U$ by $e^{i\phi/2}$
2. multiplying any single element of $U$ by $e^{i\phi}$
3. multiplying any row or column of $U$ by $e^{i\phi/2}$
4. multiplying $U$ by $e^{i\phi}$
The general $2\times 2$ unitary matrix is defined by \begin{equation*} U=\begin{pmatrix} a&b\\ -e^{i\phi}b^*&e^{i\phi}a^* \end{pmatrix} \end{equation*} Clearly, as columns or rows of $U$ form an orthonormal set, we have, $|a|^2+|b|^2=1$ and $\det{U}=e^{i\phi}$.
Let us take option (A) i.e. multiplying $U$ by $e^{-i\phi/2}$ \begin{equation*} V=e^{-i\phi}U=\begin{pmatrix} e^{-i\phi/2}a&e^{-i\phi/2}b\\ -e^{i\phi/2}b^*&e^{i\phi/2}a^* \end{pmatrix} \end{equation*} Clearly, we get, $\det{V}=1$.
Hence, option (A) is correct
4. The value of the integral $\int_C \frac{z^3dz}{z^2-5z+6}$, where $C$ is a closed contour defined by the equation $2|z|- 5 = 0$, traversed in the anti-clockwise direction, is
1. $-16\pi i$
2. $16\pi i$
3. $8\pi i$
4. $2\pi i$
\begin{align*} &\int_C\frac{z^3dz}{z^2-5z+6} \quad C:|z|=\frac{5}{2}=2.5\\ &= \int_C\frac{z^3dz}{(z-2)(z-3)}\\ &= \int_C\frac{\frac{z^3}{(z-3)}}{(z-2)}dz \end{align*} The function has a pole of order 1 at $z=2$ which lies inside $C$. According to residue theorem \begin{align*} &\oint\limits_Cf(z)dz\\ &=2\pi i\times&\text{sum of residues at}\\ &&\text{ poles inside }C \end{align*} In general the residue at a pole of order $m$ at $z=z_0$ is \begin{eqnarray} \begin{array}{c}R=\frac{1}{(m-1)!}\lim\limits_{z\to z_0}\left\{\frac{d^{m-1}\left((z-z_0)^mf(z)\right)}{dz^{m-1}}\right\}_{z=z_0}\end{array} \end{eqnarray} For simple pole of order 1 $$R=\lim\limits_{z\to z_0}\left\{(z-z_0)f(z_0)\right\}$$ $$R=\lim\limits_{z\to 2}\left\{(z-2)\frac{\frac{z^3}{(z-3)}}{(z-2)}\right\}=-8$$ $$\oint_Cf(z)dz=2\pi i\times(-8)=-16\pi i$$ Hence, answer is (A)
5. The function $f(x)$ obeys the differential equation $\frac{d^2f}{dx^2}-(3 - 2i)f = 0$ and satisfies the conditions $f(0) = 1$ and $f(x)\rightarrow \infty$ as $x\rightarrow 0$. The value of $f(\pi)$ is
1. $e^{2\pi}$
2. $e^{-2\pi}$
3. $-e^{-2\pi}$
4. $-e^{2\pi i}$

Problem set 22

1. Given the usual canonical commutation relations, the commutator $[A,B]$ of $A=i(xp_y-yp_z)$ and $B=i(yp_z+zp_y)$ is
1. $\hbar(xp_z-p_xz)$
2. $-\hbar(xp_z-p_xz)$
3. $\hbar(xp_z+p_xz)$
4. $-\hbar(xp_z+p_xz)$
Poisson bracket of two functions $f(q_i,p_i)$ and $g(q_i,p_i)$ is defined by \begin{align*} &\left\{f,g\right\}=\sum\limits_{i=1}^{N}\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\\ &=\frac{\partial f}{\partial x}\frac{\partial g}{\partial p_x}-\frac{\partial f}{\partial p_x}\frac{\partial g}{\partial q_x}+\frac{\partial f}{\partial q_y}\frac{\partial g}{\partial p_y}\\ &-\frac{\partial f}{\partial p_y}\frac{\partial g}{\partial q_y}+\frac{\partial f}{\partial q_z}\frac{\partial g}{\partial p_z}-\frac{\partial f}{\partial p_z}\frac{\partial g}{\partial q_z} \end{align*} $A=i(xp_y-yp_x)$ and $B=(yp_z+zp_y)$ \begin{align*} \left\{A,B\right\}&=ip_y\cdot 0-(-iy)\cdot 0\\ &+\:(-ip_x)\cdot z-ix\cdot p_z\\ &+\:0\cdot y-0\cdot p_y\\ &=-i(p_xz+xp_z) \end{align*} However, commutator bracket, $[q,p]=i\hbar\left\{q,p\right\}$ $$[\hat A,\hat B]=i\hbar\left\{A,B\right\}=\hbar(p_xz+xp_z)$$ Hence, option (C) is correct.
2. The entropy of a system, $S$, is related to the accessible phase space volume $\Gamma$ by $S = k\ln \Gamma(E, N,V)$ where $E$, $N$ and $V$ are the energy, number of particles and volume respectively. From this one can conclude that $\Gamma$
1. does not change during evolution to equilibrium
2. oscillates during evolution to equilibrium
3. is a maximum at equilibrium
4. is a minimum at equilibrium
Answer is (C) i.e. $\Gamma$ is a maximum at equilibrium. This is because, when system attains equilibrium its entropy and hence number of accessible states becomes maximum (equilibrium state corresponds to the most disordered state).
3. Let $\Delta W$ be the work done in a quasistatic reversible thermodynamic process. Which of the following statements about $\Delta W$ is correct?
1. $\Delta W$ is a perfect differential if the process is isothermal
2. $\Delta W$ is a perfect differential if the process is adiabatic
3. $\Delta W$ is always a perfect differential
4. $\Delta W$ cannot be a perfect differential
There are two types of functions viz. State functions and path functions. In case of state functions, the change in the state function depends only on the initial and final states of the system and not on the path from the initial to the final state (e.g. internal energy $U(S,V)$, enthalpy $H(S,P)$, Gibb's free energy $G(T,P)$ Helmholtz free energy $F(T,V)$ etc. are state functions). In case of path functions, the change in the path functions depends on the path followed from the initial to the final state (e.g..$dQ$, $dW$ etc. are path functions). State functions are exact differential while path functions are inexact differential.
According to first law of thermodynamics $$dU=dQ+dW$$ Here $dU$ is exact, but, $dQ$ and $dW$ are inexact. However, for adiabatic process $dQ=0$. Hence, we have $$dU=dW$$. As $dU$ is always exact, for adiabatic process, $dW$ is exact.
Hence answer (B) is correct.
4. Consider a system of three spins $S_1$, $S_2$ and $S_3$ each of which can take values $+1$ and $-1$. The energy of the system is given by $E = -J\left[ S_1 S_2 + S_2 S_3 + S_3 S_1\right]$, where $J$ is a positive constant. The minimum energy and the corresponding number of spin configurations are, respectively,
1. $J$ and 1
2. $-3J$ and 1
3. $-3J$ and 2
4. $-6J$ and 2
Clearly, for $S_1= S_2 = S_3=1$ or $S_1= S_2 = S_3=-1$, we get minimum energy $E=-3J$. Hence, there are 2 configurations. Hence, answer (C) is correct.
5. The minimum energy of a collection of 6 non-interacting electrons of spin-$\frac{1}{2}$ and mass $m$ placed in a one dimensional infinite square well potential of width $L$ is
1. $14\pi^2\hbar^2/mL^2$
2. $91\pi^2\hbar^2/mL^2$
3. $7\pi^2\hbar^2/mL^2$
4. $3\pi^2\hbar^2/mL^2$
For a particle in one dimensional infinite square well potential of width $L$, energy is given by $$E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\quad n=1,2,\dots$$ In each level there can be two electrons with up and down spin. Hence, total energy of the system is \begin{align*} E_{total}&=2\frac{1^2\pi^2\hbar^2}{2mL^2}+2\frac{2^2\pi^2\hbar^2}{2mL^2}\\ &+2\frac{3^2\pi^2\hbar^2}{2mL^2}\\ &=\frac{14\pi^2\hbar^2}{mL^2} \end{align*} Hence, answer is (A)

Problem set 21

1. Four charges (two $+q$ and two $-q$) are kept fixed at the four vertices of a square of side $a$ as shown
   
   At the point $P$ which is at a distance $R$ from the centre $( R > > a)$, the potential is proportional to
1. $1/R$
2. $1/R^2$
3. $1/R^3$
4. $1/R^4$
The system forms a quadrupole. The electric field and potential energy variations for various system are as follow:
For monopole $E\propto 1/r^2$ and $V\propto 1/r$
For dipole $E\propto 1/r^3$ and $V\propto 1/r^2$
For quadrupole $E\propto 1/r^4$ and $V\propto 1/r^3$
For octapole $E\propto 1/r^5$ and $V\propto 1/r^4$
Hence, answer is (C).
2. A point charge $q$ of mass $m$ is kept at a distance $d$ below a grounded infinite conducting sheet which lies in the $xy$-plane. For what value of $d$ will the charge remains stationary?
1. $q/4\sqrt{mg\pi\epsilon_0}$
2. $q/\sqrt{mg\pi\epsilon_0}$
3. There is no finite value of $d$
4. $\sqrt{mg\pi\epsilon_0}/q$

When a charge $+q$ is placed below the plate at a distance $d$, then there is an image charge $-q$ above the plate at the same distance $d$. For the charge to remain stationary, weight $mg$ must equal to electrostatic force $\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}$. i.e. $$mg=\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}$$ $$d=\frac{q}{4\sqrt{mg\pi\epsilon_0}}$$ Hence, answer is (A).
3. The wave function of a state of the hydrogen atom is given by ${\scriptstyle\Psi=\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}}$, where $\psi_{nlm}$ is the normalized eigenfunction of the state with quantum numbers $n,l$ and $m$ in the usual notation. The expectation value of $L_z$ in the state $\Psi$ is
1. $15\hbar/16$
2. $11\hbar/16$
3. $3\hbar/8$
4. $\hbar/8$
Let the normalized wavefunction is given by $$\Psi={\scriptstyle A\left[\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]}$$ Applying normalization condition \begin{align*} |\Psi|^2&=|A|^2\left[|\psi_{200}|^2+|2\psi_{211}|^2\right.\\ &\left.+|3\psi_{210}|^2+|\sqrt{2}\psi_{21-1}|^2\right]\\ &=1 \end{align*} we get, $|A|^2\left[1+4+9+2\right]=1$, hence, $A=\frac{1}{4}$. Hence, normalized wavefunction is $$\Psi={\scriptstyle \frac{1}{4}\left[\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]}$$ \begin{align*} &< L_z > =\left < \Psi|L_z|\Psi\right > \\ &=\left < \frac{1}{4}\left[\psi_{200}+2\psi_{211}+3\psi_{210}\right.\right.\\ &\left.+\sqrt{2}\psi_{21-1}\right]\left|L_z\right|\frac{1}{4}\left[\psi_{200}\right.\\ &\left.\left.+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]\right >\\ &=\frac{1}{16}\left\{\left < \psi_{200}\left|L_z\right|\psi_{200}\right >\right.\\ &+4\left < \psi_{211}\left|L_z\right|\psi_{211}\right >+9\left < \psi_{210}\left|L_z\right|\psi_{210}\right >\\ &\left.+2\left < \psi_{21-1}\left|L_z\right|\psi_{21-1}\right >\right\} \end{align*} But $L_z|\psi_{nlm}> =m\hbar|\psi_{nlm}>$ \begin{align*} < L_z >&=\frac{1}{16}\left\{0+4\hbar+0+2(-1)\hbar\right\}\\ &=\frac{1}{8}\hbar \end{align*} Hence, answer is (D)
4. The energy eigenvalues of a particle in the potential $V(x) =\frac{1}{2}m\omega^2x^2-ax$ are
1. $E_n=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{2m\omega^2}$
2. $E_n=\left(n+\frac{1}{2}\right)\hbar\omega+\frac{a^2}{2m\omega^2}$
3. $E_n=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{m\omega^2}$
4. $E_n=\left(n+\frac{1}{2}\right)\hbar\omega$
\begin{align*} V(x)&=\frac{1}{2}m\omega^2x^2-ax=\frac{1}{2}m\omega^2\left[x^2-\frac{2ax}{m\omega^2}\right]\\ &=\frac{1}{2}m\omega^2\left[x^2-\frac{2ax}{m\omega^2}+\frac{a^2}{m^2\omega^4}-\frac{a^2}{m^2\omega^4}\right]\\ &=\frac{1}{2}m\omega^2\left[\left(x-\frac{a}{m\omega^2}\right)^2-\left(\frac{a}{m\omega^2}\right)^2\right] \end{align*} Let $\left(x-\frac{a}{m\omega^2}\right)=x'$ \begin{align*} V(x')&=\frac{1}{2}m\omega^2x'^2-\frac{1}{2}m\omega^2\left(\frac{a}{m\omega^2}\right)^2\\ &=\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2} \end{align*} Second term on RHS is a constant. $$H'=-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2}$$ \begin{align*} E&=\left < \psi_n|H'|\psi_n\right > \\ &=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2}\right|\psi_n\right >\\ &=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2\right|\psi_n\right >-\left < \psi_n\left|\frac{a^2}{2m\omega^2}\right|\psi_n\right >\\ &=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2\right|\psi_n\right >-\frac{a^2}{2m\omega^2}\left < \psi_n\left.\right|\psi_n\right >\\ &=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{2m\omega^2} \end{align*} Hence, answer is (A).
5. If a particle is represented by the normalized wave function \begin{align*} \psi(x)=\begin{cases}\frac{\sqrt{15}\left(a^2-x^2\right)}{4a^{5/2}}&{\scriptstyle\text{for} -a < x < a}\\ 0 &\text{otherwise} \end{cases} \end{align*} the uncertainty $\Delta p$ in its momentum is
1. $2\hbar/5a$
2. $5\hbar/2a$
3. $\sqrt{10}\hbar/a$
4. $\sqrt{5}\hbar/\sqrt{2}a$
$$\Delta P=\sqrt{<\Delta P^2>-<\Delta P>^2}$$ \begin{align*} & <\Delta P>=\int\psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\psi\:dx\\ &=-\frac{15i\hbar}{a^516}\int\limits_{-a}^{a}(a^2-x^2)\frac{\partial(a^2-x^2)}{\partial x}dx\\ &=\frac{15i\hbar}{a^58}\int_{-a}^{a}(a^2-x^2)x\:dx=0 \end{align*} Since the integrand is an odd function. \begin{align*} & <\Delta P^2>=\int\psi^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}\right)\psi\:dx\\ &=-\frac{15\hbar^2}{a^516}\int\limits_{-a}^{a}(a^2-x^2)\frac{\partial^2(a^2-x^2)}{\partial x^2}dx\\ &=\frac{15\hbar^2}{a^58}\int_{-a}^{a}(a^2-x^2)\:dx\\ &=\frac{15\hbar^2}{a^58}\left[a^2x-\frac{x^3}{3}\right]_{-a}^{a}=\frac{5\hbar^2}{2a^2} \end{align*} $$\Delta P=\frac{\sqrt{5}\hbar}{\sqrt{2}a}$$ Hence, answer is (D).

Problem set 20

1. Let $v$, $p$ and $E$ denote the speed, the magnitude of the momentum, and the energy of a free particle of rest mass $m$. Then
1. $dE/dp=constant$
2. $p=mv$
3. $v=cp/\sqrt{p^2+m^2c^2}$
4. $E=mc^2$
As particle is moving with velocity $v$ and having rest mass $m$, its relativistic mass is given by $$m_{rel}=\frac{m}{\sqrt{1-v^2/c^2}}$$ Its momentum is given by $$p=m_{rel}v=\frac{mv}{\sqrt{1-v^2/c^2}}$$ Squaring and rearranging this equation, we get, $$p^2\left(1-v^2/c^2\right)=m^2v^2$$ On solving this equation, we get, $v=cp/\sqrt{p^2+m^2c^2}$. Hence, option $(C)$ is correct.
The rest mass energy and kinetic energy of particle are $mc^2$ and $K=\frac{mc^2}{\sqrt{1-v^2/c^2}}-mc^2$, respectively.
Here, option (D) is wrong, because, total energy $E$, is sum of kinetic energy plus rest mass energy i.e. $E=\frac{mc^2}{\sqrt{1-v^2/c^2}}$.
Option (A) is wrong, because, velocity is involved in the expression for $E$.
2. A binary star system consists of two stars $S_1$ and $S_2$, with masses $m$ and $2m$, respectively, separated by a distance $r$. If both $S_1$ and $S_2$ individually follow circular orbits around the centre of mass with instantaneous speeds $V_1$ and $V_2$ respectively, the speeds ratio $V_1/V_2$ is
1. $\sqrt{2}$
2. 1
3. 1/2
4. 2
Let the stars are at distances $r_1$ and $r_2$ from centre of mass, hence, we have $$m_1r_1=m_2r_2\quad\text{and}\quad r_1+r_2=r$$ For the motion of first star in circular orbit, we have $$\frac{Gm_1m_2}{r}=\frac{m_1v_1^2}{r_1}$$ For the motion of second star in circular orbit, we have $$\frac{Gm_1m_2}{r}=\frac{m_2v_2^2}{r_2}$$ Taking ratio of these two equations, we get, $$\frac{v_1}{v_2}=\sqrt{\frac{r_1m_2}{r_2m_1}}$$ Substituting $\frac{r_1}{r_2}=\frac{m_2}{m_1}$ $$\frac{v_1}{v_2}=\frac{m_2}{m_1}=2$$ Hence, answer is (D)
3. Three charges are located on the circumference of a circle of radius $R$ as shown in the figure below. The two charges $Q$ subtend an angle $90^o$ at the centre of the circle. The charge $q$ is symmetrically placed with respect to the charges $Q$. If the electric field at the centre of the circle is zero, what is the magnitude of $Q$?

1. $q/\sqrt{2}$
2. $\sqrt{2}q$
3. $2q$
4. $2q$

From figure, for balancing, we must have $$E_q=2E_Q\cos{(45)}$$ i.e. $\frac{1}{4\pi\epsilon_0}\frac{q}{R^2}=2\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}\frac{1}{\sqrt{2}}$. Hence, $Q=q/\sqrt{2}$.
4. Consider a hollow charged shell of inner radius $a$ and outer radius $b$. The volume charge density is $\rho(r) =\frac{k}{r^2}$ ($k$ is a constant) in the region $a < r < b$. The magnitude of the electric field produced at distance $r > a$ is
1. $\frac{k(b-a)}{\epsilon_0r^2}$ for all $r > a$
2. $\frac{k(b-a)}{\epsilon_0r^2}$ for $a < r < b$ and $\frac{kb}{\epsilon_0r^2}$ for $r > b$
3. $\frac{k(r-a)}{\epsilon_0r^2}$ for $a < r < b$ and $\frac{k(b-a)}{\epsilon_0r^2}$ for $r > b$
4. $\frac{k(r-a)}{\epsilon_0a^2}$ for $a < r < b$ and $\frac{k(b-a)}{\epsilon_0a^2}$ for $r > b$

Let us divide the region $r > a$ into two regions viz. $a\leq r\leq b$ and $r > b$.
For region $a\leq r\leq b$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ \begin{align*} E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^r\rho\:dV\\ &=\frac{1}{\epsilon_0}\int_a^r\frac{k}{r^2}\:4\pi r^2\;dr\\ &=\frac{4\pi k}{\epsilon_0}(r-a) \end{align*} $$E=\frac{k(r-a)}{\epsilon_0 r^2}$$ For region $r > b$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ \begin{align*} E\;4\pi r^2&=\frac{1}{\epsilon_0}\int_a^b\rho\:dV\\ &=\frac{1}{\epsilon_0}\int_a^b\frac{k}{r^2}\:4\pi r^2\;dr\\ &=\frac{4\pi k}{\epsilon_0}(b-a) \end{align*} $$E=\frac{k(b-a)}{\epsilon_0 r^2}$$ Hence, answer is (C)
5. Consider the interference of two coherent electromagnetic waves whose electric field vectors are given by $\vec E_1 = \hat i E_0 \cos{(\omega t+\phi)}$ and $\vec E_2 = \hat j E_0 \cos{(\omega t+\phi)}$ where $\phi$ is the phase difference. The intensity of the resulting wave is given by $\frac{\epsilon_0}{2}\left < E^2\right >$, where $\left < E^2\right > $ is the time average of $E^2$. The total intensity is
1. $0$
2. $\epsilon_0E_0^2$
3. $\epsilon_0E_0^2\sin^2{\phi}$
4. $\epsilon_0E_0^2\cos^2{\phi}$
Resultant electric field is given by \begin{align*} \vec E &= \hat i E_0 \cos{(\omega t+\phi)}\\ &+\hat j E_0 \cos{(\omega t+\phi)} \end{align*} $$ E^2 = 2 E_0^2 \cos^2{(\omega t+\phi)}$$ \begin{align*} \left < E^2\right > &=\frac{\int_0^TE^2\,dt}{\int_0^T\,dt}\\ &=\frac{2E_0^2}{T}\int_0^T\cos^2{(\omega t+\phi)}\;dt\\ &={\scriptstyle\frac{2E_0^2}{T}\frac{1}{2}\int_0^T\left[1-\cos2{(\omega t+\phi)}\right]\;dt}\\ &=\frac{E_0^2}{T}T=E_0^2 \end{align*} $$I=\frac{\epsilon_0}{2}\left < E^2\right > =\frac{\epsilon_0E_0^2}{2}$$