Problem set 64

1. Consider the differential equation $\frac{d^2x}{dt^2}-3\frac{dx}{dt}+2x=0$. If $x=0$ at $t=0$ and $x=1$ at $t=1$, the value of $x$ at $t=2$ is
1. $e^2+1$
2. $e^2+e$
3. $e+2$
4. $2e$

Characteristic equation is $$m^2-3m+2=0$$ $$\Rightarrow (m-1)(m-2)=0$$ $$\Rightarrow m=1,2$$ Hence solution is $$x=c_1e^{t}+c_2e^{2t}$$ $x=0$ at $t=0$ $\Rightarrow c_1=-c_2$

and $x=1$ at $t=1$ $\Rightarrow c_1e+c_2e^{2}=1$ $$\Rightarrow c_1=\frac{1}{e-e^2}$$ $$ c_2=-\frac{1}{e-e^2}$$ At $t=2$ \begin{align*} x&=c_1e^{2}+c_2e^{4}\\ &=\frac{e^{2}}{e-e^2}-\frac{e^{4}}{e-e^2}\\ &=e+e^2 \end{align*}

Hence, answer is (B)

2. An electric dipole of moment $\vec P$ and length $l$ aligned along the $z$-axis is used to generate electromagnetic waves. Initially it was operated at frequency 10 MHz and its power along the equatorial plane at certain distance $d (d >> l)$ was measured as $P_1$. Later, the same dipole was operated at frequency 40 MHz and power $P_2$ was measured at the same point. How do you compare the two powers ?
1. $P_2= 256 P_1$
2. $P_2= 64 P_1$
3. $P_2= 128 P_1$
4. $P_2= 16 P_1$

The average total power radiated by oscillating electric dipole is given by $$ P=\frac{\mu_0\omega^4p_0^2}{12\pi c}$$ Thus, $$ P\propto \omega^4$$ $$ P_1\propto (10)^4$$ $$ P_2\propto (40)^4$$ $$\frac{P_2}{P_1}=\left(\frac{40}{10}\right)^4$$ $$P_2= 256 P_1$$

Hence, answer is (A)

3. The number of fundamental vibrational modes of $CO_2$ molecule is :
1. 4 : 2 Raman active and 2 IR active
2. 4 : 1 Raman active and 3 IR active
3. 3 : 1 Raman active and 2 IR active
4. 3 : 2 Raman active and 1 IR active

$CO_2$ molecule is a linear molecule. Hence, number of vibrational modes are given by $3N-5$. $CO_2$ is triatomic molecule, hence $N=3$. Hence, number of modes $=3\times 3-5=4$. Out of which 1 is Raman active and 3 are IR active.

Hence, answer is (B)

4. The $L$, $S$ and $J$ quantum numbers corresponding to the ground state electronic configuration of Boron $(z = 5)$ are :
1. $L = 1$, $S = 1/2$, $J = 3/2$
2. $L = 1$, $S = 1/2$, $J = 1/2$
3. $L = 1$, $S = 3/2$, $J = 1/2$
4. $L = 0$, $S = 3/2$, $J = 3/2$

To determine ground state, we will use following Hund's rules

1. Given more than one allowed value for $S$, choose largest possible value.
2. Given more than one allowed value for $L$, choose largest possible value.
3. Given more than one allowed value for $J$, choose largest possible value of $J$ if subshell is more than half-filled and choose smallest possible value of $J$ if subshell is less than half-filled.

The ground state of Boron has a $1s^22s^22p^1$ configuration. Hence, there is single electron in $2p$ state, for which, $S=1/2$, $L=1$. Hence, $J=L-S=1/2$.

Hence, answer is (B)

5. What are the expected types of gamma ray transitions between the following states of odd ‘A’ nuclei : $$g_{9/2}\rightarrow P_{1/2}$$
1. M4 and E5
2. M1 and E2
3. M3 and E4
4. M6 and E7

Gamma rays emitted in electric $E$-$l$ transitions carry off angular momentum $l$ and parity $(-1)^l$. Therefore, angular momenta and parity of final and initial states are related by $$\Delta I=|I_i-I_f| \text{ to } I_i-I_f=l$$ $$\text{and }\Delta\pi=(-1)^l$$ The selection rules for magnetic $M$-$l$ transitions are $$\Delta I=|I_i-I_f|\text{ to } I_i-I_f=l$$ $$\text{and }\Delta\pi=(-1)^{l-1}$$ Following table shows selections rules for $\gamma$-ray transitions

Type	$l=\Delta I$	$\Delta\pi$
E1	1	yes
M1	1	no
E2	2	no
M2	2	yes
E3	3	yes
M3	3	no
E4	4	no
M4	4	yes
E5	5	yes
M5	5	no

For $g_{9/2}\rightarrow P_{1/2}$ $$\Delta I=l=\left|\frac{9}{2}-\frac{1}{2}\right|\text{ to } \frac{9}{2}+\frac{1}{2}=4, 5$$ Also, we know that the state with even $l$ value have even parity and the state with odd $l$ have odd parity. Hence, for $g_{9/2}$ parity is even and for $P_{1/2}$ parity is odd. Hence, there is parity change i.e. $$\Delta\pi=\text{ yes}$$ For $l=4,5$ and $\Delta\pi=\text{ yes}$, from above table we get that transitions as M4 and E5

Hence, answer is (A)

Problem set 63

1. Consider a system of $N$ linear polyatomic molecules. Each molecule consists of $n$ atoms. At high temperature the vibrational contribution to the specific heat is
1. $(3n-5)kN$
2. $(3n-5)\frac{kN}{2}$
3. $(3n-6)kN$
4. $(3n-6)\frac{kN}{2}$

For a polyatomic molecule containing $n$ atoms, the total number of coordinates is $3n$. Out of these, 3 coordinates are taken up for the translational motion of the molecule as a whole and 3 for rotational motion. Hence, vibrational degrees of freedom is $(3n - 6)$. However, for a linear molecule, the rotational motion along the molecular axis is not meaningful as the rotated configuration is indistinguishable from the original configuration. Therefore, for a linear molecule, there are two rotational degrees of freedom and $(3n- 5)$ vibrational degrees of freedom. According to equipartition theorem, each degree of freedom has energy $\frac{1}{2}kT$. Hence, vibrational energy consider of a linear molecule is given by $(3n-5)\frac{kT}{2}$. Hence energy of $N$ linear molecules is $$E=(3n-5)\frac{NkT}{2}$$ $$C_v=\frac{\partial E}{\partial T}=(3n-5)\frac{Nk}{2}$$

Hence, answer is (B)

2. The partition function $z(T)$ of a linear quantum mechanical harmonic oscillator in thermal equilibrium with a heat reservoir at temperature $T$ is given by:
1. $\frac{e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}$
2. $\frac{e^{-\beta\hbar\omega}}{1+e^{-\beta\hbar\omega}}$
3. $\frac{e^{-\beta\hbar\omega/2}}{1+e^{-\beta\hbar\omega}}$
4. $\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}$
where, $\hbar\omega > kT$

$$E_n=(n+\frac{1}{2})\hbar\omega$$ \begin{align*} z&=\sum\limits_{n=0}^{\infty}e^{-\beta E_n}\\ &=\sum\limits_{n=0}^{\infty}e^{-\beta (n+\frac{1}{2})\hbar\omega}\\ &=e^{ \frac{-\beta\hbar\omega}{2}}\sum\limits_{n=0}^{\infty}e^{-\beta n\hbar\omega}\\ z&=e^{ \frac{-\beta\hbar\omega}{2}}\left(1+e^{-\beta \hbar\omega}+e^{-2\beta \hbar\omega}+\cdots\right)\\ \end{align*} This is a geometric series where each term is obtained from previous term by multiplying $e^{-\beta \hbar\omega}$. Hence, $$\left(1+e^{-\beta \hbar\omega}+e^{-2\beta \hbar\omega}+..\right)\!\!=\!\!\frac{1}{1-e^{-\beta\hbar\omega}}$$ $$z=\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}$$

Hence, answer is (D)

3. The output of a laser has a pulse width of 30 ms and average output power of 0.6 watt per pulse. If the wavelength of the laser light is 640 nm. How many photon does each pulse contain?
1. $2.9\times10^{18}$
2. $3.5\times10^{18}$
3. $5.8\times10^{15}$
4. $6.5\times10^{16}$

$\Delta t=30 ms=30\times10^{-3}s$, $P_{avg}=0.6 \text{watt per pulse}$, $\lambda=640 nm=640\times10^{-9}m$ Energy per pulse is power times pulse duration. \begin{align*} E_{pulse}&=P_{avg}\Delta t\\ &=0.6\times30\times10^{-3}\\ &=0.018 joule \end{align*} Energy per photon is given by Planck formula. \begin{align*} E_{photon}&=h\nu=\frac{hc}{\lambda}\\ &=\frac{6.63\times10^{-34}\times3\times10^8}{640\times10^{-9}}\\ &=0.031078125\times10^{-17} joule \end{align*} Number of photons per pulse is just energy per pulse divided by energy per photon \begin{align*} N&=\frac{E_{pulse}}{E_{photon}}\\ &=\frac{0.018}{0.031078125\times10^{-17}}\\ &=5.8\times10^{16} \end{align*}

Hence, answer is (C)

4. In a band structure calculation, the dispersion relation for electrons is found to be $$\epsilon_k = \beta( \cos{k_xa} + \cos{k_ya} +\cos{k_za}),$$ where $\beta$ is a constant and $a$ is the lattice constant. The effective mass at the boundary of the first Brillouin zone is
1. $\frac{2\hbar^2}{5\beta a^2}$
2. $\frac{4\hbar^2}{5\beta a^2}$
3. $\frac{\hbar^2}{2\beta a^2}$
4. $\frac{\hbar^2}{3\beta a^2}$

$$m^*=\frac{\hbar^2}{\frac{\partial^2\epsilon_k}{\partial k^2}}$$ At first Brillouin zone boundary $(k_x,k_y,k_z)=\left(\frac{\pi}{a},\frac{\pi}{a},\frac{\pi}{a}\right)$. Let us consider only first term in energy i.e. $$\epsilon_{k_x} = \beta( \cos{k_xa})$$ $$\frac{\partial\epsilon_{k_x}}{\partial k}=\beta\left( -a\sin{k_xa}\frac{\partial k_x}{\partial k}\right)$$ Using $k=\sqrt{k_x^2+k_y^2+k_z^2}$ $$\frac{\partial k_x}{\partial k}=\frac{ k}{ k_x}$$ $$\frac{\partial\epsilon_{k_x}}{\partial k}=\beta\left( -a\frac{ k}{ k_x}\sin{k_xa}\right)$$ \begin{align*} \frac{\partial^2\epsilon_{k_x}}{\partial k^2}&=\beta\left( -a\frac{ 1}{ k_x}\sin{k_xa}\right.\\ &~~~-ak\sin{k_xa}\frac{ \partial}{\partial k} \frac{1}{k_x}\\ &\left.~~~-a^2\frac{ k^2}{ k_x^2}\cos{k_xa}\right) \end{align*} \begin{align*} \frac{\partial^2\epsilon_{k_x}}{\partial k^2}\left|_{k_x,k_y,k_z=\frac{\pi}{a}}\right.&=\beta\left( a^2\frac{ k^2}{ k_x^2}\right)\\ &=\beta a^2\left( \frac{ 3\frac{ \pi^2}{ a^2}}{ \frac{ \pi^2}{ a^2}}\right)\\ &=3\beta a^2 \end{align*} Similarly, $$\frac{\partial^2\epsilon_{k_y}}{\partial k^2}=3\beta a^2$$ $$\frac{\partial^2\epsilon_{k_z}}{\partial k^2}=3\beta a^2$$ $$m^*=\frac{\hbar^2}{3\beta a^2}$$

Hence, answer is (D)

5. Consider the energy level diagram shown below, which corresponds to the molecular nitrogen laser.
   
   If the pump rate $R$ is $10^{20}$ atoms cm$^{-3}$ s$^{-1}$ and the decay routes are as shown with $\tau_{21} =20\: ns$ and. $\tau_{1} = 1\: \mu s$, the equilibrium populations of states 2 and 1 are, respectively,
1. $10^{14}$ cm$^{-3}$ and $2\times10^{12}$ cm$^{-3}$
2. $2\times10^{12}$ cm$^{-3}$ and $10^{14}$ cm$^{-3}$
3. $2\times10^{12}$ cm$^{-3}$ and $2\times10^{6}$ cm$^{-3}$
4. zero and $10^{20}$ cm$^{-3}$

$$\frac{dN_2}{dt}=R-\frac{N_2}{\tau_{21}}$$ and $$\frac{dN_1}{dt}=\frac{N_2}{\tau_{21}}-\frac{N_1}{\tau_{1}}$$ At equilibrium $\frac{dN_2}{dt}=\frac{dN_1}{dt}=0$ $$N_2=\tau_{21}R=2\times10^{12}\: cm^{-3}$$ $$N_1=\frac{\tau_1N_2}{\tau_{21}}=10^{14}\:cm^{-3}$$

Hence, answer is (B)

Problem set 62

1. The energy levels of one-dimensional harmonic oscillator with potential $V(x)=\frac{1}{2}kx^2$ are given by $h\nu\left(n+\frac{1}{2}\right)$ with $n=0,1,2,\dots$. If the potential is changed to $V(x)=\infty$ for $x<0$ and $V(x)=\frac{1}{2}kx^2$ for $x>0$, the energy levels now, will be given by:
1. $h\nu\left(n+\frac{3}{2}\right)$
2. $2h\nu\left(n+\frac{1}{2}\right)$
3. $h\nu\left(n+\frac{1}{2}\right)$, $n$ odd only
4. $h\nu\left(n+\frac{1}{2}\right)$, $n$ even only

The wave function of simple harmonic oscillator with potential $V(x)=\frac{1}{2}kx^2$ is given by $$\psi_n(x)=N_nH_n\left(\sqrt{\alpha}x\right)e^{-\alpha x^2/2}$$ where $\alpha=\left(\frac{mk}{\hbar^2}\right)^{1/2}$ is constant and $N_n=\left(\frac{\alpha}{\pi}\right)^{1/4}\left(\frac{1}{2^nn!}\right)^{1/2}$ is normalization constant, and $H_n\left(\sqrt{\alpha}x\right)=H_n(z)$ is a Hermite polynomial of degree $n$ given by $$H_0(z)=1$$ $$H_1(z)=2z$$ $$H_2(z)=4z^2-2$$ $$H_3(z)=8z^3-12z$$ The energy eigenvalues are given by \begin{align*} E_n&=h\nu\left(n+\frac{1}{2}\right)\\ &=\frac{1}{2}h\nu,\frac{3}{2}h\nu,\frac{5}{2}h\nu\dots\\ &\text{for }n=0,1,2,3\dots \end{align*} Now, as $V(x)=\infty$ for $x<0$, the wave functions must vanish at $x=0$. Clearly, one can see that the wavefunction vanish at $x=0$ only for odd values of $n$. The terms with even $n$ do not have node at $x = 0$. Hence, eneegy is given by \begin{align*} E_n&=h\nu\left(n+\frac{1}{2}\right)\\ &=\frac{3}{2}h\nu,\frac{7}{2}h\nu,\frac{11}{2}h\nu\dots\\ & \text{for }n=1,3,5\dots \end{align*}

Hence, answer is (C)

2. At $t=0$ a one-dimensional harmonic oscillator is in a state given by:$$\Psi(x,0)=\frac{1}{2}u_0(x)+i\frac{\sqrt{3}}{2}u_1(x)$$ where $u_0$ and $u_1$ are first two normalized eigen states. ($\omega$ is natural angular frequency of the oscillator). Then
1. The expectation value of the energy is $\frac{5}{4}\hbar\omega$
2. Energy measurement of this state will always gives its value as $\frac{5}{4}\hbar\omega$
3. The average value of energy is $\hbar\omega$
4. The expectation value of energy is dependent on time

Clearly, one can verify that the wavefunction $\Psi$ is normalized. \begin{align*} & < E > = < \Psi|H|\Psi > \\ &={\scriptstyle\left < \frac{1}{2}u_0+i\frac{\sqrt{3}}{2}u_1\arrowvert H\arrowvert\frac{1}{2}u_0+i\frac{\sqrt{3}}{2}u_1\right > }\\ &=\left < \frac{1}{2}u_0\left|H\right|\frac{1}{2}u_0\right > \\ &+\left < i\frac{\sqrt{3}}{2}u_1\left|H\right|i\frac{\sqrt{3}}{2}u_1\right > \\ &=\frac{1}{4}\left < u_0\left|H\right|u_0\right >\\ & +\!\!\left(\frac{-i\sqrt{3}}{2}\right)\!\!\left(\frac{i\sqrt{3}}{2}\right)\!\!\left < u_1\left|H\right|u_1\right > \\ &=\frac{1}{4}E_0+\frac{3}{4}E_1\\ &=\frac{1}{4}\times\frac{1}{2}\hbar\omega+\frac{3}{4}\times\frac{3}{2}\hbar\omega\\ < E >&=\frac{5}{4}\hbar\omega \end{align*}

Hence, answer is (A)

3. If the $\phi$ dependent part of the eigen function of an electron in a Hydrogen atom is $e^{2i\phi}$, then the minimum principal and minimum orbital angular momentum quantum numbers $n$ and $l$ respectively for this eigen function will be:
1. $n=3$, $l=2$
2. $n=2$, $l=1$
3. $n=1$, $l=2$
4. $n=2$, $l=2$

The $\phi$ dependent part of the eigen function of an electron in a Hydrogen atom is given by $e^{im\phi}$. Hence, $m=2$. Now, for a given value of $l$, $m$ has values from $-l$ to $+l$. This implies that $l=2$. Also, for a given value of $n$, $l$ takes values from $0$ to $n-1$. Hence, for $l=2$, minimum value of $n$ should be $3$.

Hence, answer is (A)

4. The ground state energy of a particle in an infinite square well is 1eV. If four particles obeying Bose-Einstein statistics are kept in this well, then the ground state energy will be :
1. 30 eV
2. 10 eV
3. 4 eV
4. $\frac{1}{4}$ eV

As particles obey Bose-Einstein statistics, all particles will be the ground state. Hence, energy will be 4 eV.

Hence, answer is (C)

5. Consider a system in contact with a heat and particle reservoir. It may be unoccupied or occupied by one particle with energy 0 and $\epsilon$. The grand partition function will be ($\beta=1/kT$)
1. $Z(\mu,T)=e^{-\epsilon\beta}$
2. $Z(\mu,T)=\left(1+e^{-\epsilon\beta}\right)^{-1}$
3. $Z(\mu,T)=1+e^{-\epsilon\beta}$
4. $Z(\mu,T)=1+e^{\mu\beta}+e^{(\mu-\epsilon)\beta}$

$$Z(\mu,T)=\sum\limits_{i} e^{-\beta n_i(\epsilon_i-\mu)}$$ where $i$ runs over every microstate, $n_i$ number particles occupying the microstate $i$, $\epsilon_i$ is energy of a particles in the $i^{th}$ microstate

Let us write all microstates

Microstate no. $i$	State with enrgy 0	State with enrgy $\epsilon$	energy of microstate $\epsilon_i$	$n_i$
1	0	0	0	0
2	1	0	0	1
3	0	1	$\epsilon$	1

\begin{align*} Z(\mu,T)&= e^{-\beta \times0\times(0-\mu)}+e^{-\beta \times1\times(0-\mu)}\\ &+e^{-\beta\times 1\times(\epsilon-\mu)} \end{align*} $$Z(\mu,T)=1+e^{\mu\beta}+e^{(\mu-\epsilon)\beta}$$

Hence, answer is (D)

Problem set 61

1. The product of the uncertainties $\left(\Delta L_x\right)\left(\Delta L_y\right)$ for a particle in the state $a|1,1 > +b|1,-1 > $ (where $|l,m > $ denotes an eigenstate of $L^2$ and $L_z$) will be a minimum for
1. $a=\pm ib$
2. $a=0$ and $b=1$
3. $a=\frac{\sqrt{3}}{2}$ and $b=\frac{1}{2}$
4. $a=\pm b$

\begin{align*} L_z |l,m >&=m\hbar|l,m >\\ L^2 |l,m >&=m(m+1)\hbar|l,m >\\ L_+ |l,m >&={\scriptstyle\sqrt{(l-m)(l+m+1)}\hbar|l,m+1 >}\\ L_- |l,m >&={\scriptstyle\sqrt{(l+m)(l-m+1)}\hbar|l,m-1 >} \end{align*} Let $|\psi >=a|1,1 > +b|1,-1 > $ \begin{align*} L_z |\psi >&=a\hbar |1,1 >-b\hbar |1,-1 > \\ L^2 |\psi >&=2a\hbar |1,1 > \\ L_+ |\psi >&=\sqrt{2}b\hbar |1,0 > \\ L_- |\psi >&=\sqrt{2}a\hbar |1,0 > \\ L_+^2 |\psi >&=2b\hbar^2 |1,1 > \\ L_-^2 |\psi >&=2a\hbar^2 |1,-1 > \end{align*} $$\left(\Delta L_x\right)=\sqrt{< L_x^2 > -< L_x >^2 }$$ \begin{align*} &< L_x >=\left <\psi|L_x|\psi\right>\\ &=\frac{1}{2}\left <\psi|L_++L_-|\psi\right>\\ &=\frac{1}{2}\left <\psi|L_+|\psi\right>+\frac{1}{2}\left <\psi|L_-|\psi\right>\\ &=\frac{1}{2}\sqrt{2}b\left <\psi|1,0\right>+\frac{1}{2}\sqrt{2}a\left <\psi|1,0\right>\\ &=0 \end{align*} \begin{align*} &< L_x^2 >=\left <\psi|L_x^2|\psi\right>\\ &=\frac{1}{4}\left <\psi|(L_++L_-)(L_++L_-)|\psi\right>\\ &={\scriptstyle\frac{1}{4}\left <\psi|L_+^2+L_-^2+L_+L_-+L_-L_+|\psi\right>}\\ &={\scriptstyle\frac{1}{4}\left[2a^*b\hbar^2+2ab^*\hbar^2+2\hbar^2|a|^2+2\hbar^2|b|^2\right]}\\ &=\frac{\hbar^2}{2}\left[a^*b+ab^*+|a|^2+|b|^2\right]\\ &=\frac{\hbar^2}{2}\left[a^*b+ab^*+1\right] \end{align*} $$\left(\Delta L_x\right)=\sqrt{\frac{\hbar^2}{2}\left[a^*b+ab^*+1\right]}$$ Similarly $$\left(\Delta L_y\right)=\sqrt{\frac{\hbar^2}{2}\left[1-(a^*b+ab^*)\right]}$$ $$\left(\Delta L_x\right)\!\left(\Delta L_y\right)=\!\frac{\hbar^2}{2}\sqrt{1-(a^*b+ab^*)^2}$$ Clearly, for $a=\pm b$, we get,$\left(\Delta L_x\right)\left(\Delta L_y\right)=0$

Hence, answer is (D)

2. Of the nuclei of mass number $A=125$, the binding energy calculated from the liquid drop model (given that the coefficients for the Coulomb and the asymmetry energy are $a_c=0.7$ MeV and $a_{sym}=22.5$ MeV respectively) is a maximum for
1. $^{125}_{54}$Xe
2. $^{125}_{53}$I
3. $^{125}_{52}$Te
4. $^{125}_{51}$Sb

Using relation $$Z=\frac{4a_{sym}+a_cA^{-1/3}}{8a_{sym}A^{-1}+2a_cA^{-1/3}}$$ we get $Z=52$

Hence, answer is (C)

3. The value of $\oint\limits_C\frac{e^{2z}}{(z+1)^4}dz$, where $C$ is circle defined by $|z|=3$, is
1. $\frac{8\pi i}{3}e^{-2}$
2. $\frac{8\pi i}{3}e^{-1}$
3. $\frac{8\pi i}{3}e$
4. $\frac{8\pi i}{3}e^{2}$

Inside circle $|z|=3$, $z=-1$ is a pole of order 4. \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} $$R=\frac{1}{3!}\lim\limits_{z\to -1}\left\{\frac{d^3\left((z+1)^4e^{2z}\right)}{dz^3}\right\}$$ $$R=\frac{8}{6}e^{-2}$$ $$\oint_Cf(z)dz=2\pi i\times\frac{8}{6}e^{-2}=\frac{8\pi i}{3}e^{-2} $$

Hence, answer is (A)

4. Consider the following processes involving free particles
   1. $\bar n\rightarrow \bar p+e^++\bar \nu_e$
   2. $\bar p+n\rightarrow \pi^-$
   3. $ p+n\rightarrow \pi^++\pi^0+\pi^0$
   4. $p+\bar \nu_e\rightarrow n+e^+$
   Which of the following statements is true?
1. Process (i) obeys all conservation laws
2. Process (ii) conserves baryon number, but violates energy-momentum conservation
3. Process (iii) is not allowed by strong interactions, but is allowed by weak interactions
4. Process (iv) conserves baryon number, but violates lepton number conservation

	$\bar n\rightarrow \bar p+e^++\bar \nu_e$
Q	$0\rightarrow -1+1+0$	Conserved
B	$-1\rightarrow -1+0+0$	Conserved
L	$0\rightarrow 0-1-1$	Not Conserved
$I_3$	$-\frac{1}{2}\rightarrow-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}$	Not Conserved

Similarly, one can check for other process. Process (ii) conserves baryon number, but violates energy-momentum conservation

Hence, answer is (B)

5. A one-dimensional harmonic oscillator is in the state $\psi(x)=\frac{1}{\sqrt{14}}\left[3\psi_0(x)-2\psi_1(x)-\psi_2(x)\right]$, where $\psi_0(x)$, $\psi_1(x)$ and $\psi_2(x)$ are the ground, first and second excited states respectively. The probability of finding the oscillator in the ground state is
1. 0
2. $\frac{3}{\sqrt{14}}$
3. $\frac{9}{14}$
4. 1

Any state of the harmonic oscillator can be expressed in terms of eigen states as follow $$\sum\limits_nc_n\psi_n(x)$$ Probability of finding the system in state $\psi_n(x)$ is given by $|c_n|^2$. Hence The probability of finding the oscillator in the ground state is $\frac{9}{14}$

Hence, answer is (C)

Problem set 60

1. A beam of unpolarized light in a medium with dielectric constant $\epsilon_1$ is reflected from a plane interface formed with another medium of dielectric constant $\epsilon_2=3\epsilon_1$. The two media have identical magnetic permeability. If the angle of incidence is $60^0$, then the reflected light
1. is plane polarized perpendicular to the plane of incidence
2. is plane polarized parallel to the plane of incidence
3. is circularly polarized
4. has the same polarization as the incident light

This is a simple method (discovered by malus in 1808) of obtaining partially plane polarized light by reflection. The percentage of polarized light is greatest in reflected beam when light beam is incident on the transparent medium with an incident angle equal to the angle of polarization. The vibrations of this plane polarized reflected light are found to be perpendicular to the plane of incidence and therefore, the reflected light is said to be plane polarized in the plane of incidence, when the incident angle is equal to the Brewster angle $\theta_B$, where $\tan\theta_B = n_2/n_1$. \begin{align*} \tan\theta_B&=\frac{n_2}{n_1}\\ &=\sqrt{\frac{\epsilon_2}{\epsilon_1}}\\ \tan\theta_B&=\sqrt{3} \end{align*} Hence, $\theta_B=60^0$. As angle of incidence is equal to Brewster angle, reflected light is plane polarized perpendicular to the plane of incidence.

Hence, answer is (A)

2. The partition function of a system of $N$ Ising spins is $Z=\lambda_1^N+\lambda_2^N$, where $\lambda_1$ and $\lambda_2$ are functions of temperature, but are independent of $N$. If $\lambda_1 > \lambda_2$, the free energy per spin in the limit $N\rightarrow\infty$ is
1. $-k_BT\ln{\left(\frac{\lambda_1}{\lambda_2}\right)}$
2. $-k_BT\ln{\lambda_2}$
3. $-k_BT\ln{\left(\lambda_1\lambda_2\right)}$
4. $-k_BT\ln{\lambda_1}$

\begin{align*} F&=-kT\ln{Z}\\ &=-kT\ln{\left(\lambda_1^N+\lambda_2^N\right)}\\ F&=-kT\ln{\left(\lambda_1^N\left[1+\left(\frac{\lambda_2}{\lambda_1}\right)^N\right]\right)}\\ \end{align*} For $\lambda_1 > \lambda_2$, $\frac{\lambda_2}{\lambda_1} < 1$, in the limit $N\rightarrow\infty$, $\frac{\lambda_2}{\lambda_1} = 0$ $$F=-kT\ln{\left(\lambda_1^N\right)}=-NkT\ln{\lambda_1^N}$$ $$\frac{F}{N}=-kT\ln{\lambda_1}$$

Hence, answer is (D)

3. The Hamiltonian of a system of $N$ non-interacting spin-1/2 particles is $H=-\mu_0B\sum_iS_i^z$, where $S_i^z=\pm1$ are the components of $i^{th}$ spin along an external magnetic field $B$. At a temperature $T$ such that $e^{\mu_0B/k_BT}=2$, the specific heat per particle is
1. $\frac{16}{25}k_B$
2. $\frac{8}{25}k_B\ln2$
3. $k_B\left(\ln2\right)^2$
4. $\frac{16}{25}k_B\left(\ln2\right)^2$

Because each spin is independent of the others and distinguishable, we can find the partition function for one spin, $Z_1$, and use the relation $Z_N=Z_1^N$ to obtain, the partition function for $N$ spins. \begin{align*} Z_1&=\sum\limits_{s=\pm1}e^{-\beta\mu_0 Bs}\\ &=e^{-\beta\mu_0 B}+e^{\beta\mu_0 B}\\ &=2\cosh{\beta\mu_0 B} \end{align*} $$Z_N=\left(2\cosh{\beta\mu_0 B}\right)^N$$ Mean energy \begin{align*} E&=-\frac{\partial\ln{Z_N}}{\partial\beta}\\ &=-N\mu B\tanh{\beta\mu_0 B} \end{align*} \begin{align*} C_V&=\frac{\partial E}{\partial T}\\ &=\left(\frac{\mu_0 B}{k T}\right)^2Nk\:sech^2{\beta\mu_0 B} \end{align*} $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{4}{\left(e^{\frac{\mu_0 B}{kT}}+e^{-\frac{\mu_0 B}{kT}}\right)^2}$$ Since, $e^{\mu_0B/k_BT}=2$ $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{4}{\left(2+\frac{1}{2}\right)^2}$$ $$\frac{C_v}{N}=\left(\frac{\mu_0 B}{k T}\right)^2k\frac{16}{25}$$ $$\frac{C_v}{N}=\left(\ln2\right)^2k\frac{16}{25}$$

Hence, answer is (D)

4. The ground state energy of a particle in the potential $V(x)=g|x|$, estimated using the trial wavefunction $$\psi(x)= \begin{cases} \sqrt{\frac{c}{a^5}}(a^2-x^2),\quad x < |a|\\ 0,\quad x\ge|a| \end{cases} $$ (where $g$ and $c$ are constants) is
1. $\frac{15}{16}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
2. $\frac{5}{6}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
3. $\frac{3}{4}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$
4. $\frac{7}{8}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$

First normalize the wavefunction using normalization condition $\int\limits_{-a}^{a}\psi^*\psi\:dx=1$ $$\frac{c}{a^5}\int\limits_{-a}^{a}(a^2-x^2)^2\:dx=1$$ $$\Rightarrow c=\frac{15}{16}$$ $$\psi(x)=\sqrt{\frac{15}{16a^5}}(a^2-x^2)$$ $$H=-\frac{\hbar^2}{2m}\frac{d^2}{d x^2}+g|x|=T+V$$ \begin{align*} < E >& = < \psi |H| \psi > \\ &= < \psi |T| \psi>+ < \psi |V| \psi> \end{align*} \begin{align*} < \psi |T| \psi> &={\scriptstyle-\frac{\hbar^2}{2m}\frac{15}{16a^5}\int\limits_{-a}^{a}(a^2-x^2) \frac{d^2(a^2-x^2)}{d x^2}dx}\\ &=\frac{\hbar^2}{m}\frac{15}{16a^5}\int\limits_{-a}^{a}(a^2-x^2)dx\\ &=\frac{5\hbar^2}{4ma^2} \end{align*} \begin{align*} < \psi |V| \psi>\! &=\frac{15}{16a^5}g\!\!\int\limits_{-a}^{a}\!(a^2-x^2)^2 |x|dx\\ &=\frac{15}{16a^5}2g\!\!\int\limits_{0}^{a}\!(a^2-x^2)^2 x dx\\ &=\frac{5}{16}ga \end{align*} $$ < E > =\frac{5\hbar^2}{4ma^2}+\frac{5}{16}ga$$ $$\frac{d < E > }{da}=0$$ $$\Rightarrow a=2\left(\frac{\hbar^2}{mg}\right)^{1/3}$$ $$ < E > =\frac{15}{16}\left(\frac{\hbar^2g^2}{m}\right)^{1/3}$$

Hence, answer is (A)

5. A hydrogen atom is subjected to the perturbation $V_{pert}(\vec r)=\epsilon\cos{\left(2r/a_0\right)}$, where $a_0$ is the Bohr radius. The change in the ground state energy to first order in $\epsilon$ is
1. $\epsilon/4$
2. $\epsilon/2$
3. $-\epsilon/2$
4. $-\epsilon/4$

First order correction is given by $$E_0^{(1)}=\left < \psi_{100}\left|H'\right|\psi_{100}\right >$$ $$\psi_{100}=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}$$ \begin{align*} E_0^{(1)}&=\frac{\epsilon}{\pi a_0^3}\!\int\limits_0^\infty\! e^{-\frac{2r}{a_0}}\!\cos{\left(\frac{2r}{a_0}\right)}4\pi r^2 dr\\ &={\textstyle\frac{4\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}}\left(\frac{e^{i\frac{2r}{a_0}}+e^{-i\frac{2r}{a_0}} }{2}\right) r^2 dr}\\ &=\frac{2\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}(1-i)}r^2dr\\ &+\frac{2\epsilon}{a_0^3}\int\limits_0^\infty\! e^{-\frac{2r}{a_0}(1+i)}r^2dr\\ &={\textstyle\frac{2\epsilon}{a_0^3}\left[\frac{2!}{\left(\frac{2}{a_0}(1-i)\right)^3}+\frac{2!}{\left(\frac{2}{a_0}(1+i)\right)^3}\right]}\\ &=\frac{\epsilon}{2}\left[\frac{1}{(1-i)^3}+\frac{1}{(1+i)^3}\right]\\ &={\textstyle\frac{\epsilon}{2}\left[\frac{1}{\left(\sqrt{2}\right)^3\left(\frac{1-i}{\sqrt{2}}\right)^3}+\frac{1}{\left(\sqrt{2}\right)^3\left(\frac{1+i}{\sqrt{2}}\right)^3}\right]}\\ &=\frac{\epsilon}{4\sqrt{2}}\left[\frac{1}{e^{\frac{-i3\pi}{4}}}+\frac{1}{e^{\frac{i3\pi}{4}}}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[e^{\frac{i3\pi}{4}}+e^{\frac{-i3\pi}{4}}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[2\cos{\left(\frac{3\pi}{4}\right)}\right]\\ &=\frac{\epsilon}{4\sqrt{2}}\left[2\left(-\frac{1}{\sqrt{2}}\right)\right]\\ &=-\frac{\epsilon}{4} \end{align*}

Hence, answer is (D)

Problem set 59

1. A dipole of moment $\vec p$, oscillating at frequency $\omega$, radiates spherical waves. The vector potential at large distance is $$\vec A(\vec r)=\frac{\mu_0}{4\pi}i\omega\frac{e^{ikr}}{r}\vec p$$. To order $(1/r)$ the magnetic field $\vec B$ at a point $\vec r=r\hat n$ is
1. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\left(\hat n\cdot\vec p\right)\frac{e^{ikr}}{r}$
2. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\left(\hat n\times\vec p\right)\frac{e^{ikr}}{r}$
3. $-\frac{\mu_0}{4\pi}\omega^2k\left(\hat n\cdot\vec p\right)\vec p\frac{e^{ikr}}{r}$
4. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\vec p\frac{e^{ikr}}{r}$

Since $\vec B=\vec\nabla\times\vec A$, option (B) is correct

Hence, answer is (B)

2. For a two level system, the population of atoms in the upper and lower levels are $3\times10^{18}$ and $0.7\times10^{18}$, respectively. If the coefficient of stimulated emission is $3\times10^{5}\:m^3/W-s^3$ and the energy density is $9.0 J/m^3Hz$, the rate of stimulated emission will be
1. $6.3\times10^{16}\:s^{-1}$
2. $4.1\times10^{16}\:s^{-1}$
3. $2.7\times10^{16}\:s^{-1}$
4. $1.8\times10^{16}\:s^{-1}$

Rate of stimulated emission is $$R_{stimu}=\rho B$$ $$R_{stimu}= 9.0 \times 3\times10^{5}$$ $$R_{stimu}=2.7\times10^{16}\:s^{-1}$$

Hence, answer is (C)

3. The first ionization potential of K is 4.34 eV, the electron affinity of Cl is 3.82 eV and the equilibrium separation of KCl is 0.3 nm. The energy required to dissociate a KCl molecule into a K and a Cl atom is
1. 8.62 eV
2. 8.16 eV
3. 4.28 eV
4. 4.14 eV

The ionization energy to form K is 4.34 eV, hence we will require 4.34 eV energy to form $K^+$. Now, the electron affinity of Cl is 3.82 eV. Hence, when the electron liberated from potassium is put on Cl to make $Cl^-$ 3.82 eV energy liberated. Hence, net amount of energy to produce $K^+$ and $Cl^-$ is 4.34-3.82=0.52 eV.

The energy required to dissociate KCl into $K^+$ and $Cl^-$ is nothing but the potential energy of attraction between $K^+$ and $Cl^-$ ions, given by \begin{align*} V&=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}\\ &={\scriptstyle-9\times10^9\times\frac{\left(1.6\times10^{-19}\right)^2}{0.3\times10^{-9}}}\\ &=-7.7\times10^{-19}\:J\\ &=-4.79\:eV \end{align*} Thus, the energy required to dissociate KCl into $K^+$ and $Cl^-$ is $4.79\:eV$

Hence, The energy required to dissociate KCl into K and Cl is $4.79-0.52=4.27\:eV$

Hence, answer is (C)

4. Considers circuits as shown in figures (a) and (b) below.
   
   
   If transistors in figures (a) and (b) have current gain ($\beta_{dc}$) of 100 and 10 respectively, then they operate in the
1. active region and saturation region respectively
2. saturation region and active region respectively
3. saturation region in both cases
4. active region in both cases

For figure (a) $I_B=\frac{10.7-0.7}{10k}=1\:mA$. Hence, $I_c=\beta I_B=100\:mA$. Thus, \begin{align*} V_{CB}&=V_C-V_B\\ &=(10-2\times100)-0.7\\ &=-ve \end{align*}

Hence, collector to base junction is forward biased.

As both base to emitter and collector to base junctions are forward biased, the transistor is in saturation region.

For figure (b)$I_B=\frac{5-0.7}{10k}=0.43\:mA$. Hence, $I_c=\beta I_B=4.3\:mA$. Thus, \begin{align*} V_{CB}&=V_C-V_B\\ &=(10-1\times4.3)-0.7\\ &=+ve \end{align*}

Hence, collector to base junction is reverse biased.

As base to emitter junction is forward biased and collector to base junctions is reversed biased, the transistor is in active region.

Hence, answer is (B)

5. A small magnetic needle is kept at $(0,0)$ with its moment along the x-axis. Another small magnetic needle is at the point $(1, 1)$ and is free to rotate in the $xy$-plane. In equilibrium the angle $\theta$ between their magnetic moments is such that
1. $\tan\theta=1/3$
2. $\tan\theta=0$
3. $\tan\theta=3$
4. $\tan\theta=1$

Potential energy of two dipole system is given by \begin{align*} U&=\frac{\mu_0}{4\pi r^3}\left[\vec m_1\cdot\vec m_2-3\vec m_1\cdot\vec r)(\vec m_2\cdot\vec r) \right]\\ &={\scriptstyle\frac{\mu_0m_1m_2}{4\pi r^3}\left[\cos\theta -3(\cos{45^0})(\cos{\left(\theta-45^0\right)}) \right]}\\ \end{align*} For equilibrium potential energy is minimum i.e. $\frac{\partial U}{\partial\theta}=0$ \begin{align*} \frac{\partial U}{\partial\theta}&={\scriptstyle\frac{\mu_0m_1m_2}{4\pi r^3}\left[\sin\theta +\frac{\sqrt{3}}{2}\sin{\left(\theta-45^0\right)} \right]}\\ &=0\\ \end{align*} $$\Rightarrow \tan\theta=3$$

Hence, answer is (C)

Problem set 58

1. For an electron moving through a one- dimensional periodic lattice of periodicity a, which of the following corresponds to an energy eigenfunction consistent with Bloch’s theorem?
1. $\psi(x)=A\exp{\left( i\left[\frac{\pi x}{a}+\cos{\left(\frac{\pi x}{2a}\right)}\right]\right)}$
2. ${\scriptstyle\psi(x)=A\exp{\left( i\left[\frac{\pi x}{a}+\cos{\left(\frac{2\pi x}{a}\right)}\right]\right)}}$
3. ${\scriptstyle \psi(x)=A\exp{\left( i\left[\frac{2\pi x}{a}+i\cosh{\left(\frac{2\pi x}{a}\right)}\right]\right)}}$
4. $\psi(x)=A\exp{\left( i\left[\frac{\pi x}{2a}+i\left|\frac{\pi x}{2a}\right|\right]\right)}$

According to Bloch theorem $$\psi(x)=e^{-ikx}u(x)$$ where $\psi(x)$ and $u(x)$ have same periods. Clearly, $\psi(x)=A\exp{\left( i\left[\frac{\pi x}{a}+\cos{\left(\frac{2\pi x}{a}\right)}\right]\right)}$ satisfies above condition.

Hence, answer is (B)

2. A thin metal film of dimension $2\:mm\times2\:mm$ contains $4\times10^{12}$ electrons. The magnitude of the Fermi wavevector of the system, in the free electron approximation, is
1. $2\sqrt{\pi}\times10^{7}\:cm^{-1}$
2. $\sqrt{2\pi}\times10^{7}\:cm^{-1}$
3. $\sqrt{\pi}\times10^{7}\:cm^{-1}$
4. $2\pi\times10^{7}\:cm^{-1}$

In two-dimensional electron gas approximation magnitude of Fermi wave vector is given by $$k^2=2\pi n$$ where, $n$ is electron density. \begin{align*} k&=\sqrt{2\pi n}\\ &=\sqrt{2\pi}\sqrt{\frac{N}{A}}\\ &=\sqrt{2\pi}\sqrt{\frac{4\times10^{12}}{4\times 10^{-2}}}\\ &=\sqrt{2\pi}\times10^{7}\:cm^{-1} \end{align*}

Hence, answer is (B)

3. A positron is suddenly absorbed by the nucleus of a tritium $(^3_1H)$ atom to turn the latter into a $He^+$ ion. If the electron in the tritium atom was initially in the ground state, the probability that the resulting $He^+$ ion will be in its ground state is
1. 1
2. $\frac{8}{9}$
3. $\frac{128}{243}$
4. $\frac{512}{729}$

Both Tritium and $He^+$ are hydrogenic atoms. The ground state wavefunction of hydrogenic atom is given by $$\psi_{0}=R_{10}Y_{00}$$ where $$R_{10}=2\left(\frac{Z}{a_0}\right)^{3/2}e^{-Zr/a_0}$$ $$Y_{00}=\frac{1}{\sqrt{4\pi}}$$ $$a_0=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}$$ For both Tritium and $He^+$ have reduced mass $\mu$ same. Hence, these two differs in $Z$ values. $$\psi_{0_{He^+}}=2\frac{1}{\sqrt{4\pi}}\left(\frac{2}{a_0}\right)^{3/2}e^{-2r/a_0}$$ $$\psi_{0_{Tritium}}=2\frac{1}{\sqrt{4\pi}}\left(\frac{1}{a_0}\right)^{3/2}e^{-r/a_0}$$ According to sudden approximation probability of transition is given by $$P=\left|\left < \psi_{0_{He^+}}| \psi_{0_{Tritium}}\right >\right|^2$$ \begin{align*} &\left < \psi_{0_{He^+}}| \psi_{0_{Tritium}}\right >\\ &={\textstyle 2^2\frac{1}{4\pi}\left(\frac{1}{a_0}\right)^32^{3/2}\int_0^\infty e^{-3r/a_0}4\pi r^2 dr}\\ &={\textstyle \left(\frac{1}{a_0}\right)^32^{7/2}\int_0^\infty r^2 e^{-3r/a_0} dr}\\ &={\scriptstyle \left(\frac{1}{a_0}\right)^32^{7/2}\!\left(\frac{a_0}{3}\right)^3\int_0^\infty \left(\frac{3}{a_0}r\right)^2 e^{-3r/a_0} d\left(\frac{3}{a_0}r\right)}\\ &={\textstyle \left(\frac{1}{a_0}\right)^32^{7/2}\left(\frac{a_0}{3}\right)^32!}\\ &=\frac{2^{9/2}}{27} \end{align*} where we have used formula $\int_0^\infty x^n e^{-x} dx=n!$ $$P=\frac{2^{9}}{27^2}=\frac{512}{729}$$

Hence, answer is (D)

4. The first order diffraction peak of a crystalline solid occurs at a scattering angle of $30^0$ when the diffraction pattern is recorded using an x-ray beam of wavelength 0.15 nm. If the error in measurements of the wavelength and the angle are 0.01 nm and $1^0$ respectively, then the error in calculating the inter- planar spacing will approximately be
1. $1.1\times 10^{-2}\:nm$
2. $1.3\times 10^{-4}\:nm$
3. $2.5\times 10^{-2}\:nm$
4. $2.0\times 10^{-3}\:nm$

$$2d\sin\theta=\lambda$$ $$d=\frac{\lambda}{2\sin\theta}$$ Let $z=\sin\theta$. $$d=\frac{\lambda}{2z}$$ \begin{align*} \Delta z&=\cos\theta\Delta\theta\\ &=\sin{30}\times 1^0\\ &=\frac{\sqrt{3}}{2}\frac{\pi}{180} \end{align*} \begin{align*} \Delta d&=\left|\frac{\lambda}{2z}\right|\sqrt{\left(\frac{\Delta\lambda}{\lambda}\right)^2+\left(\frac{\Delta z}{z}\right)^2}\\ &={\textstyle\left|\frac{0.15}{2\sin{30}}\right|\sqrt{\left(\frac{0.01}{0.15}\right)^2+\left(\frac{\frac{\sqrt{3}}{2}\frac{\pi}{180}}{\sin{30}}\right)^2}}\\ &=\left|0.15\right|\sqrt{\frac{1}{225}+\frac{3\pi^2}{180^2}}\\ &=0.011=1.1\times 10^{-2}\:nm \end{align*}

Hence, answer is (A)

5. The dispersion relation of electrons in a 3-dimensional lattice in the tight binding approximation is given by, $$\epsilon_k=\alpha\cos{k_xa}+\beta\cos{k_ya}+\gamma\cos{k_za}$$ where $a$ is the lattice constant and $\alpha,\beta,\gamma$ are constants with dimension of energy. The effective mass tensor at the corner of the first Brillouin zone $\left(\frac{\pi}{a},\frac{\pi}{a},\frac{\pi}{a}\right)$ is
1. $\frac{\hbar^2}{a^2}\begin{pmatrix}-\frac{1}{\alpha}&0&0\\0&-\frac{1}{\beta}&0\\0&0&\frac{1}{\gamma}\end{pmatrix}$
2. $\frac{\hbar^2}{a^2}\begin{pmatrix}-\frac{1}{\alpha}&0&0\\0&-\frac{1}{\beta}&0\\0&0&-\frac{1}{\gamma}\end{pmatrix}$
3. $\frac{\hbar^2}{a^2}\begin{pmatrix}\frac{1}{\alpha}&0&0\\0&\frac{1}{\beta}&0\\0&0&\frac{1}{\gamma}\end{pmatrix}$
4. $\frac{\hbar^2}{a^2}\begin{pmatrix}\frac{1}{\alpha}&0&0\\0&\frac{1}{\beta}&0\\0&0&-\frac{1}{\gamma}\end{pmatrix}$

Effective mass tensor is given by $$m^*=\begin{pmatrix}m_{xx}&m_{xy}&m_{xz}\\m_{yx}&m_{yy}&m_{yz}\\m_{zx}&m_{zy}&m_{zz}\end{pmatrix}$$ where $$m_{\alpha\beta}=\frac{\hbar^2}{\frac{\partial^2\epsilon_k}{\partial k_\alpha \partial k_\beta}}\quad \alpha ,\beta=x,y,z$$ $$m_{xx}=\frac{\hbar^2}{\frac{\partial^2\epsilon_k}{\partial k_x \partial k_x}}=-\frac{\hbar^2}{a^2}\frac{1}{\alpha}\cos{k_xa}$$ $$\left.m_{xx}\right|_{\frac{\pi}{a}}=\frac{\hbar^2}{a^2}\frac{1}{\alpha}$$ Similarly, find other

Hence, answer is (C)

Problem set 57

1. Three variables $a$, $b$, $c$ are each randomly chosen from uniform distribution in the interval $[0,1]$. The probability that $a+b>2c$ is
1. $\frac{3}{4}$
2. $\frac{2}{3}$
3. $\frac{1}{2}$
4. $\frac{1}{4}$

The condition $a+b>2c$ can be written as $\frac{a+b}{2}>c$. Here, $\frac{a+b}{2}$ is average value $a$ and $b$. Hence, $\frac{a+b}{2}$ lies in the interval $[0,1]$. Also, $c$ lies in the interval $[0,1]$. Hence, probability that $\frac{a+b}{2} > c$ is $\frac{1}{2}$.

Hence, answer is (C)

2. Which one of the following sets of Maxwell's equations for time independent charge density $\rho$ and current density $\vec J$ is correct?
1. \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=-\frac{\partial\vec B}{\partial t}\\ \vec\nabla\times\vec B&=\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*}
2. \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}
3. \begin{align*} \vec\nabla\cdot\vec E&=0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}
4. \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=\mu_0|\vec J |\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*}

Maxwell equations are given by \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=-\frac{\partial\vec B}{\partial t}\\ \vec\nabla\times\vec B&=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial\vec E}{\partial t} \end{align*} For time independent case $\frac{\partial\vec B}{\partial t}=0$ and $\frac{\partial\vec E}{\partial t}=0$. Hence, Maxwell equations become \begin{align*} \vec\nabla\cdot\vec E&=\rho/\epsilon_0\\ \vec\nabla\cdot\vec B&=0\\ \vec\nabla\times\vec E&=0\\ \vec\nabla\times\vec B&=\mu_0\vec J \end{align*}

Hence, answer is (B)

3. A system of $N$ non-interacting classical particles, each of mass $m$ is in a two-dimensional harmonic potential of the form $V(r)=\alpha(x^2+y^2)$ where $\alpha$ is a positive constant. The canonical partition function of the system at temperature $T$ $\left(\beta=\frac{1}{k_BT}\right)$:
1. $\left[\left(\frac{\alpha}{2m}\right)^2\frac{\pi}{\beta}\right]^N$
2. $\left(\frac{2m\pi}{\alpha\beta}\right)^{2N}$
3. $\left(\frac{\alpha\pi}{2m\beta}\right)^N$
4. $\left(\frac{2m\pi^2}{\alpha\beta^2}\right)^{N}$

The classical partition is given by $$Z=\frac{1}{h^s}\int\cdots\int e^{-\beta H(q,p)}\:dq\:dp$$ OR $$Z=\int\cdots\int e^{-\beta H(q,p)}\:dq\:dp$$ In second definition the factor $\frac{1}{h^s}$ is missing. Here, $s$ is dimensionality. This factor arises when treating the classical result as a limiting scenario of a quantum mechanical problem. In a purely classical context, however, this factor should not be included.

For 2D harmonic oscillator $$H=\frac{p_x^2}{2m}+\frac{p_x^2}{2m}+\alpha(x^2+y^2)$$ Hence, single particle canonical partition function is given by \begin{align*} Z_1&=\!\!\int\!\!\!\int\! e^{-\beta \left(\!\frac{p_x^2}{2m}\!+\!\frac{p_y^2}{2m}\!+\!\alpha(x^2\!+\!y^2)\!\right)}\!dxdydp_xdp_y\\ &=\int\limits_{-\infty}^{\infty}e^{-\beta \frac{p_x^2}{2m}}\:dp_x\int\limits_{-\infty}^{\infty}e^{-\beta \frac{p_y^2}{2m}}\:dp_y\\ &\int\limits_{-\infty}^{\infty} e^{-\beta \alpha x^2}\:dx\int\limits_{-\infty}^{\infty} e^{-\beta \alpha y^2}\:dy\\ &=\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{\pi }{\alpha\beta}}\sqrt{\frac{\pi }{\alpha\beta}}\\ &=\frac{2\pi^2 m}{\alpha\beta^2} \end{align*} Here, we have used formula $\int\limits_{-\infty}^{\infty} e^{-a x^2}\:dx=\sqrt{\frac{\pi }{a}}$

For $N$ particle system the partition function is given by $$Z=Z_1^N=\left(\frac{2\pi^2 m}{\alpha\beta^2}\right)^N$$

Hence, answer is (D)

4. In a two-state system, the transition rate of a particle from state 1 to state 2 is $t_{12}$, and the transition rate of a particle from state 2 to state 1 is $t_{21}$. In the steady state, the probability of finding the particle in state 1 is
1. $\frac{t_{21}}{t_{12}+t_{21}}$
2. $\frac{t_{12}}{t_{12}+t_{21}}$
3. $\frac{t_{12}t_{21}}{t_{12}+t_{21}}$
4. $\frac{t_{12}-t_{21}}{t_{12}+t_{21}}$

Let $P_1$ be the probability of transition from state 1 to state 2 and $P_2$ be the probability of transition from state 2 to state 1. Then the rate equations can be written as $$\frac{d}{dt}\begin{pmatrix}P_1\\P_2\end{pmatrix}=\begin{pmatrix}-t_{21}&t_{12}\\t_{21}&-t_{12}\end{pmatrix}\begin{pmatrix}P_1\\P_2\end{pmatrix}$$ In steady state, we have,$$\frac{d}{dt}\begin{pmatrix}P_1\\P_2\end{pmatrix}=0$$ Hence, $$\begin{pmatrix}-t_{21}&t_{12}\\t_{21}&-t_{12}\end{pmatrix}\begin{pmatrix}P_1\\P_2\end{pmatrix}=0$$ This gives, $$t_{21}P_1=t_{12}P_2$$ Hence, $$P_1=\frac{t_{12}}{t_{21}}P_2$$ $$P_1=\frac{t_{12}}{t_{21}}(1-P_1)$$ $$P_1\left(1+\frac{t_{12}}{t_{21}}\right)=\frac{t_{12}}{t_{21}}$$ $$P_1=\frac{t_{21}}{t_{12}+t_{21}}$$

Hence, answer is (A)

5. The viscosity $\eta$ of a liquid is given by Poiseuille's formula $\eta=\frac{\pi Pa^4}{8lV}$. Assume that $l$ and $V$ can be measured very accurately, but the pressure $P$ has an rms error of 1% and the radius $a$ has an independent rms error of 3%. The rms error of viscosity is closest to
1. 2%
2. 4%
3. 12%
4. 13%

If $\Delta x$ and $\Delta y$ are rms errors in $x$ and $y$ then error in $z=x^my^n$ is given by $$\frac{\Delta z}{z}=\sqrt{\left(\frac{m\Delta x}{x}\right)^2+\left(\frac{n\Delta y}{y}\right)^2}$$ Let us take $\frac{\pi }{8lV}=1$. Hence, $\eta=Pa^4$. $$\Delta \eta=\eta\sqrt{\left(\frac{\Delta P}{P}\right)^2+\left(\frac{4\Delta a}{a}\right)^2}$$ Let $P=1$ and $a=1$. Hence, $\eta=1$ $$\Delta \eta=\sqrt{\left(\Delta P\right)^2+\left(4\Delta a\right)^2}$$ $$\Delta \eta=\sqrt{\left(\frac{1}{100}\right)^2+\left(4\times\frac{3}{100}\right)^2}$$ $$\Delta \eta=\frac{1}{100}\sqrt{145}$$ $$\Delta \eta=\frac{12.04}{100}$$ $$\Delta \eta=12.04\%$$

Hence, answer is (C)

Problem set 56

1. A system of $N$ distinguishable particles, each of which can be in one of the two energy levels $0$ and $\epsilon$, has a total energy $n\epsilon$, where $n$ is an integer. The entropy of the system is proportional to
1. $N \ln{n}$
2. $n \ln{N}$
3. $\ln{\frac{N!}{n!}}$
4. $\ln{\left(\frac{N!}{n!(N-n)!}\right)}$

Let us consider three distinguishable particles A, B, C to be arranged in two energy levels $0$ and $\epsilon$, such that total energy is $n\epsilon$ ($n=0,1,3$). The total number of microstates are given as follow:

Total energy $n\epsilon$	0	$\epsilon$	Number of microstates
$0\epsilon$	ABC	-	1
$1\epsilon$	AB	C	3
	AC	B
	BC	A
$2\epsilon$	A	BC	3
	B	AC
	C	AB
$3\epsilon$	-	ABC	1

In general, the number microstates for $N$ particles for total energy $n\epsilon$ is given by $$\Omega=\frac{N!}{n!(N-n)!}$$ Hence, entropy of system is given by $$S=k_B\ln\Omega=k_B\ln{\left(\frac{N!}{n!(N-n)!}\right)}$$

Hence, answer is (D)

2. The wavefunction of a particle in one-dimension is denoted by $\psi(x)$ in the coordinate representation and $\phi(p)=\int \psi(x) e^{-ipx/\hbar}\:dx$ in the momentum representation. If the action of an operator $\hat T$ on $\psi(x)$ is given by $\hat T\psi(x)=\psi(x+a)$, where $a$ is constant, then $\hat T\phi(p)$ is given by
1. $-\frac{i}{\hbar}ap\phi(p)$
2. $e^{-iap/\hbar}\phi(p)$
3. $e^{+iap/\hbar}\phi(p)$
4. $\left(1+\frac{i}{\hbar}ap\right)\phi(p)$

$$\phi(p)=\int \psi(x) e^{-ipx/\hbar}\:dx$$ \begin{align*} \hat T\phi(p)&=\int \hat T\psi(x) e^{-ipx/\hbar}\:dx\\ &=\int \psi(x+a) e^{-ipx/\hbar}\:dx\\ &=\!\int\!\psi(x+a) e^{\frac{-ip(x+a)}{\hbar}}e^{\frac{+ipa}{\hbar}}dx\\ &=e^{\frac{+ipa}{\hbar}}\int\! \psi(x+a) e^{\frac{-ip(x+a)}{\hbar}}dx\\ &=e^{+ipa/\hbar}\phi(p) \end{align*}

Hence, answer is (C)

3. A proton moves with a speed of 300m/s in a circular orbit in the xy-plane in a magnetic 1 tesla along the positive z-direction. When an electric field of 1 V/m is applied along the positive y-direction, the center of the circular orbit
1. remains stationary
2. moves at 1 m/s along the negative x-direction
3. moves at 1 m/s along the positive z-direction
4. moves at 1 m/s along the positive x-direction

The addition of an electric field perpendicular to a given magnetic field simply causes the particle to drift perpendicular to both the electric and magnetic field with the fixed velocity $$\vec v_{EB}=\frac{\vec E\times\vec B}{B^2}$$ In the present case, $\vec E=1\hat j$ and $\vec B=1\hat k$ $$\vec v_{EB}=\frac{\hat j\times\hat k}{1}=\hat i$$ Hence, center of circular moves at 1 m/s along the positive x-direction

Hence, answer is (D)

4. Suppose the yz-plane forms a chargeless boundary between two media of permittivities $\epsilon_{left}$ and $\epsilon_{right}$ where $\epsilon_{left}:\epsilon_{right}=1:2$. If the uniform electric field on the left is $\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$ (where $c$ is constant), then the electric field on the right $\vec E_{right}$ is
1. $c\left(2\hat i+\hat j+\hat k\right)$
2. $c\left(\hat i+2\hat j+2\hat k\right)$
3. $c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$
4. $c\left(\hat i+\frac{1}{2}\hat j+\frac{1}{2}\hat k\right)$

Here the electric fields in two media should satisfy the condition $$ E_{left}\epsilon_{left}\cos\theta_1= E_{right}\epsilon_{right}\cos\theta_2$$ where, $\theta_1$ is the angle between $\vec E_{left}$ and unit normal to yz-plane in left medium and $\theta_2$ is the angle between $\vec E_{right}$ and unit normal to yz-plane in right medium. $$\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$$ The unit vector normal to yz-plane in left medium is $\hat n=-\hat i$. Hence, angle between $\vec E_{left}$ and $\hat n$ is given by \begin{align*} \cos\theta_1&=\frac{\left|\vec E_{left}\cdot\hat n\right|}{\left|\vec E_{left}\right|\left|\hat n\right|}\\ &=\frac{c}{c\sqrt{3}\times 1}\\ \cos\theta_1&=\frac{1}{\sqrt{3}} \end{align*} \begin{align*} E_{right}\cos\theta_2&= E_{left}\frac{\epsilon_{left}}{\epsilon_{right}}\cos\theta_1\\ &=c\sqrt{3}\times\frac{1}{2}\times\frac{1}{\sqrt{3}}\\ E_{right}\cos\theta_2&=\frac{c}{2} \end{align*} The vector which satisfies above condition is $\vec E_{right}=c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$

Hence, answer is (C)

5. A particle of mass $2\: kg$ is moving such that at time $t$, its position, in metre, is given by $\vec r(t) = 5\hat i- 2t^2\hat j$. The angular momentum of the particle at $t = 2\: s$ about the origin, in $kg\: m^2\: s^{-1}$, is
1. $-40\hat k$
2. $-80\hat k$
3. $80\hat k$
4. $40\hat k$

$$\vec r(t) = 5\hat i- 2t^2\hat j$$ $$\vec v(t) = - 4t\hat j$$ \begin{align*} \vec L&=m\left(\vec r\times\vec v\right)\\ &=m\begin{vmatrix}\hat i&\hat j&\hat k\\5&- 2t^2&0\\0&-4t&0\end{vmatrix}\\ &=-20mt\hat k=-80\hat k \end{align*}

Hence, answer is (B)

Problem set 55

1. The mass $m$ of a moving particle is $\frac{2m_0}{\sqrt{3}}$, where $m_0$ is its rest mass. The linear momentum of the particle is
1. $2m_0c$
2. $\frac{2m_0c}{\sqrt{3}}$
3. $m_0c$
4. $\frac{m_0c}{\sqrt{3}}$

Given $m=\frac{2m_0}{\sqrt{3}}$. Relativistic mass is given by $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ Hence, $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{2m_0}{\sqrt{3}}$$ Hence, $$\sqrt{1-\frac{v^2}{c^2}}=\frac{\sqrt{3}}{2}$$ $$1-\frac{v^2}{c^2}=\frac{3}{4}$$ $$\frac{v^2}{c^2}=1-\frac{3}{4}=\frac{1}{4}$$ $$v=\frac{c}{2}$$ Momentum is given by \begin{align*} p&=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}\\ &=\frac{2m_0}{\sqrt{3}}\frac{c}{2}\\ &=\frac{m_0c}{\sqrt{3}} \end{align*}

Hence, answer is (D)

2. For the given transformations (i) $Q = p$, $P = -q$ and (ii) $Q = p$, $P = q$, where $p$ and $q$ are canonically conjugate variables, which one of the following statements is true?
1. Both (i) and (ii) are canonical
2. Only (i) is canonical
3. Only (ii) is canonical
4. Neither (i) nor (ii) is canonical
A transformation is said be canonical if it satisfies Poisson Bracket $\left\{Q,P\right\}_{q,p}=1$ i.e. $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=1$$ For (i) $Q = p$, $P = -q$ $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=0-1\times-1=1$$ Hence, transformation is canonical

For (ii) $Q = p$, $P = q$ $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=0-1\times1=-1$$ Hence, transformation is not canonical

Hence, answer is (B)

3. Consider three inertial frames of reference A, B and C. The frame B moves with a velocity $c/2$ with respect to $A$, and $C$ moves with velocity $c/10$ with respect to B in the same direction. The velocity of C as measured in A is
1. $\frac{3c}{7}$
2. $\frac{4c}{7}$
3. $\frac{c}{7}$
4. $\frac{\sqrt{3}c}{7}$

Let $V$ be the velocity of frame $B$ with respect to frame A, $u_x$ be the velocity of frame C with respect to frame A, and $u_x'$ be the velocity of frame C with respect to frame B, then we have $${\textstyle u_x'=\frac{u_x-V}{1-u_xV/c^2}\quad\text{and}\quad u_x=\frac{u_x'+V}{1+u_x'V/c^2}}$$ \begin{align*} u_x&=\frac{u_x'+V}{1+u_x'V/c^2}\\ &=\frac{\frac{c}{10}+\frac{c}{2}}{1+\frac{c}{10}\frac{c}{2}\frac{1}{c^2}}\\ &=\frac{4}{7}c \end{align*}

Hence, answer is (B)

4. A system of four particles is in $x$-$y$ plane. Of these, two particles each of mass $m$ are located at $( -1, 1)$ and $(1, -1)$. The remaining two particles each of mass $2m$ are located at $(1, 1)$ and $( -1, -1)$. The $xy$-component of the moment of inertia tensor of this system of particles is
1. $10m$
2. $-10m$
3. $2m$
4. $-2m$

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)\\ &=m+m+2m+2m\\ &=6m \end{align*} \begin{align*} &I_{12}= -\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\ &={\scriptscriptstyle -(m(-1)(1)+m(1)(-1)+2m(1)(1)+2m(-1)(-1))}\\ &=-(-m-m+2m+2m)\\ &=-2m \end{align*} \begin{align*} &I_{21}= -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}\\ &={\scriptscriptstyle -(m(1)(-1)+m(-1)(1)+2m(1)(1)+2m(-1)(-1))}\\ &=-2m \end{align*} \begin{align*} I_{22}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right)\\ &=m+m+2m+2m\\ &=6m \end{align*} $${I}=\begin{pmatrix} 6m&-2m\\ -2m&6m \end{pmatrix} $$ Hence, $xy$-component of moment of inertia is $-2m$

Hence, answer is (D)

5. A particle of mass $m$ is represented by the wavefunction $\psi(x) = Ae^{ikx}$, where $k$ is the wavevector and $A$ is a constant. The magnitude of the probability current density of the particle is
1. $|A|^2\frac{\hbar k}{m}$
2. $|A|^2\frac{\hbar k}{2m}$
3. $|A|^2\frac{\left(\hbar k\right)^2}{m}$
4. $|A|^2\frac{\left(\hbar k\right)^2}{2m}$

$$j=\frac{i\hbar}{2m}\left(\psi\frac{\partial\psi^*}{\partial x}-\psi^*\frac{\partial\psi}{\partial x}\right)$$ $$\psi(x) = Ae^{ikx}$$ $$\psi^*(x) = A^*e^{-ikx}$$ $$\frac{\partial\psi}{\partial x} = ikAe^{ikx}$$ $$\frac{\partial\psi^*}{\partial x} = -ikA^*e^{-ikx}$$ $$\psi^*\frac{\partial\psi}{\partial x} = A^*e^{-ikx} ikAe^{ikx}=ik|A|^2$$ \begin{align*} \psi\frac{\partial\psi^*}{\partial x} &= Ae^{ikx}\left(-ikA^*e^{-ikx}\right)\\ &=-ik|A|^2 \end{align*} \begin{align*} j&=\frac{i\hbar}{2m}\left(-ik|A|^2-ik|A|^2 \right)\\ &=|A|^2\frac{\hbar k}{m} \end{align*}

Hence, answer is (A)

Problem set 54

1. The high input impedance of field effect transistor (FET) amplifier is due to
1. the pinch-off voltage
2. its very low gate current
3. the source and drain being far apart
4. the geometry of the FET

The pinch-off voltage

Hence, answer is (A)

2. The circuit shown in the figure functions as
   
1. an OR gate
2. an AND gate
3. a NOR gate
4. a NAND gate

When either or both inputs A and B will be high, the respective transistors will be made ON and output will be high, hence circuit functions as OR gate

Hence, answer is (A)

3. A linear transformation $T$, defined as $T\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}x_1+x_2\\x_2-x_3\end{pmatrix}$, transforms a vector $\vec x$ from a three-dimensional real space to a two-dimensional real space. The transformation matrix $T$ is
1. $\begin{pmatrix}1&1&0\\0&1&-1\end{pmatrix}$
2. $\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}$
3. $\begin{pmatrix}1&1&1\\-1&1&1\end{pmatrix}$
4. $\begin{pmatrix}1&0&0\\0&0&1\end{pmatrix}$

\begin{align*} T\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}&=\begin{pmatrix}1&1&0\\0&1&-1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\\ &=\begin{pmatrix}x_1+x_2\\x_2-x_3\end{pmatrix} \end{align*}

Hence, answer is (A)

4. Three point charges $q$, $q$ and $-2q$ are located at $(0,-a,a)$, $(O,a,a)$ and $(0,0,-a)$, respectively. The net dipole moment of this charge distribution is
1. $4qa\hat k$
2. $2qa\hat k$
3. $-4qa\hat i$
4. $-2qa\hat j$

If the system of charges is electrically neutral as whole, then the net dipole moment of a system of charges $q_1,q_2,\dots, q_n$ with position vectors $r_1,r_2,\dots, r_n$ is given by $$\vec P=\sum\limits_{i=1}^{n}q_i\vec r_i$$ \begin{align*} \vec P&=q\vec r_1+q\vec r_2-2q\vec r_3\\ &=q\left(-a\hat j+a\hat k\right)+q\left(a\hat j+a\hat k\right)-2q\left(-a\hat k\right)\\ &=4qa\hat k \end{align*}

Hence, answer is (A)

5. The vector potential in a region is given as $\vec A(x, y, z) = -y\hat i + 2x\hat j$. The associated magnetic induction $\vec B$ is
1. $\hat i+\hat k$
2. $3\hat k$
3. $-\hat i+2\hat j$
4. $-\hat i+\hat j+\hat k$

\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}\\ -y&2x&0\end{vmatrix}\\ &=3\hat k \end{align*}

Hence, answer is (B)

Problem set 53

1. Which one of the following axes of rotational symmetry is NOT permissible in single crystals?
1. two-fold axis
2. three-fold axis
3. four-fold axis
4. five-fold axis

The rotational symmetries of a crystal are usually limited to 2-fold, 3-fold, 4-fold, and 6-fold.

Hence, answer is (D)

2. Weak nuclear forces act on
1. both hadrons and leptons
2. hadrons only
3. all particles
4. all charged particles

Weak nuclear forces act on both hadrons and leptons, while strong nuclear force acts on hadrons only.

Hence, answer is (A)

3. Which one of the following disintegration series of the heavy elements will give $^{209}Bi$ as a stable nucleus?
1. Thorium series
2. Neptunium series
3. Uranium series
4. Actinium series

Uranium series terminates with lead-206

Actinium series terminates with lead-207

Thorium series terminates with lead-208

Neptunium series terminates with bismuth-209 and thallium-205

Hence, answer is (B)

4. The order of magnitude of the binding energy per nucleon in a nucleus is
1. $10^{-5}$ MeV
2. $10^{-3}$ MeV
3. 0.1 MeV
4. 10 MeV

Hence, answer is (D)

5. The interaction potential between two quarks, separated by a distance $r$ inside a nucleon, can be described by ($a$, $b$ and $\beta$ are positive constants)
1. $ae^{-\beta r}$
2. $\frac{a}{r}+br$
3. $-\frac{a}{r}+br$
4. $\frac{a}{r}$

The interaction potential between two quarks, separated by a distance $r$ inside a nucleon, can be described by $\frac{a}{r}+br$

Hence, answer is (B)

Problem set 52

1. Which one of the following relations is true for Pauli matrices $\sigma_x$, $\sigma_y$ and $\sigma_z$
1. $\sigma_x\sigma_y=\sigma_y\sigma_x$
2. $\sigma_x\sigma_y=\sigma_z$
3. $\sigma_x\sigma_y=i\sigma_z$
4. $\sigma_x\sigma_y=-\sigma_y\sigma_x$

Pauli spin matrices are $\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\sigma_y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$, and $\sigma_z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ \begin{align*} \sigma_x\sigma_y&=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}0&-i\\i&0\end{pmatrix}\\ &=\begin{pmatrix}i&0\\0&-i\end{pmatrix}\\ &=i\sigma_z \end{align*}

Hence, answer is (C)

2. The free energy of a photon gas enclosed in a volume $V$ is given by $F=-\frac{1}{3}aVT^4$, where $a$ is constant and $T$ is the temperature of the gas. The chemical potential of the photon gas is
1. 0
2. $\frac{4}{3}aVT^4$
3. $\frac{1}{3}aT^4$
4. $aVT^4$

Chemical potential is given by $$\mu=\frac{\partial F}{\partial N}$$ $$\mu=\frac{\partial (-\frac{1}{3}aVT^4)}{\partial N}=0$$

Hence, answer is (A)

3. The wavefunctions of two identical particles in states $n$ and $s$ are given by $\phi_n(r_1)$ and $\phi_s(r_2)$, respectively. The particles obey Maxwell-Boltzmann statistics. The state of the combined two-particle system is expressed as
1. $\phi_n(r_1)+\phi_s(r_2)$
2. $\frac{1}{\sqrt{2}}\left[\phi_n(r_1)\phi_s(r_2)+\phi_n(r_2)\phi_s(r_1)\right]$
3. $\frac{1}{\sqrt{2}}\left[\phi_n(r_1)\phi_s(r_2)-\phi_n(r_2)\phi_s(r_1)\right]$
4. $\phi_n(r_1)\phi_s(r_2)$

In Maxwell-Boltzmann statistics particles are distinguishable, hence, $$\Phi=\phi_n(r_1)\phi_s(r_2)$$ or $$\Phi=\phi_n(r_2)\phi_s(r_1)$$

Hence, answer is (D)

4. The target of an X-ray tube is subjected to an excitation voltage $V$. The wavelength of the emitted X-rays is proportional to
1. $1/\sqrt{V}$
2. $\sqrt{V}$
3. $1/V$
4. $V$

Energy of electrons is $$E_k=eV=h\nu$$ $$\lambda=\frac{hc}{eV}$$

Hence, answer is (C)

5. Which one of the following is NOT a correct statement about semiconductors?
1. The electrons and holes have different mobilities in a semiconductor
2. In $n$-type semiconductor, the Fermi level lies closer to the conduction band edge
3. Silicon is a direct band gap semiconductor
4. Silicon has diamond structure

Silicon is a indirect band gap semiconductor

Hence, answer is (C)

Problem set 51

1. The trace of $3\times3$ matrix is 2. Two of its eigenvalues are 1 and 2. The third eigenvalue is
1. -1
2. 0
3. 1
4. 2

The trace of a matrix is the sum of the eigenvalues. Hence, $1+2+x=2$, hence $x=-1$

Hence, answer is (A)

2. A particle is moving in an inverse square force field. If the total energy of the particle is positive, the trajectory of the particle is
1. circular
2. elliptical
3. parabolic
4. hyperbolic

Orbit	Eccentricity $\epsilon$	Energy $E$
Circular	$\epsilon=0$	$E < 0$
Parabolic	$\epsilon=1$	$E=0$
Elliptical	$0 < \epsilon < 1$	$E < 0$
Hyperbolic	$\epsilon > 1$	$E\ge0$

Hence, answer is (D)

3. In an electromagnetic field, which one of the following remains invariant under Lorentz transformation?
1. $\vec E\times\vec B$
2. $E^2-c^2B^2$
3. $B^2$
4. $E^2$

The Lorentz transformation equations of electric and magnetic field are given by

$E_x'=E_x$	$B_x'=B_x$
${\scriptstyle E_y'=\gamma\left(E_y-vB_z\right)}$	${\scriptstyle B_y'=\gamma\left(B_y+\frac{v}{c^2}E_z\right)}$
${\scriptstyle E_z'=\gamma\left(E_z+vB_y\right)}$	${\scriptstyle B_z'=\gamma\left(B_z-\frac{v}{c^2}E_y\right)}$

\begin{align*} &E^{2'}-c^2B^{2'}= E_x^{2'}+E_y^{2'}+E_z^{2'}\\ &-c^2\left(B_x^{2'}+B_y^{2'}+B_z^{2'}\right)\\ &={\textstyle E_x^2+\left(\gamma\left(E_y-vB_z\right)\right)^2}\\ &{\textstyle +\left(\gamma\left(E_z+vB_y\right)\right)^2}\\ &-c^2{\scriptstyle \left(B_x^2+\left(\gamma\left(B_y+\frac{v}{c^2}E_z\right) \right)^2+\left(\gamma\left(B_z-\frac{v}{c^2}E_y\right)\right)^2 \right)}\\ &=E^2-c^2B^2 \end{align*}

Hence, answer is (B)

4. A sphere of radius $R$ has uniform volume charge density. The electric potential at a point $r(r < R)$ is
1. due to the charge inside a sphere of radius $r$ only
2. due to the entire charge of the sphere
3. due to the charge of the spherical shell of inner and outer radii $r$ and $R$, only
4. independent of $r$

For region $r < R$, using Gauss law, we have, $$\int\vec E\cdot d\vec s=\frac{q_{enclosed}}{\epsilon_0}$$ $$E4\pi r^2=\frac{\frac{4}{3}\pi r^3\rho}{\epsilon_0}$$ $$E=\frac{\rho}{3\epsilon_0}r$$ \begin{align*} V&=-\int\limits_0^r\frac{\rho}{3\epsilon_0}r\:dr\\ &=-\frac{\rho}{6\epsilon_0}r^2 \end{align*}

Hence, answer is (A)

5. A free particle is moving in $+x$ direction with a linear momentum $p$. The wavefunction of the particle normalized in a length $L$ is
1. $\frac{1}{\sqrt{L}}\sin{\frac{p}{\hbar}x}$
2. $\frac{1}{\sqrt{L}}\cos{\frac{p}{\hbar}x}$
3. $\frac{1}{\sqrt{L}}e^{-i\frac{p}{\hbar}x}$
4. $\frac{1}{\sqrt{L}}e^{i\frac{p}{\hbar}x}$

For a free particle potential $V=0$, hence, for a one dimensional motion, Schrodinger equation is $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$$ $$i.e. \frac{d^2\psi}{dx^2}=-\frac{2mE}{\hbar^2}\psi$$ $$or \frac{d^2\psi}{dx^2}=-k^2\psi$$ where, $k^2=\frac{2mE}{\hbar^2}$. The possible solutions of the above equation are \begin{align*} \psi(x)=A \begin{cases} \sin{kx}=\sin{\sqrt{\frac{2mE}{\hbar^2}}x}\\ \cos{kx}=\cos{\sqrt{\frac{2mE}{\hbar^2}}x}\\ e^{\pm ikx}=e^{\pm i\sqrt{\frac{2mE}{\hbar^2}}x} \end{cases} \end{align*} where $A$ is normalization constant. For a particle moving in $+x$ direction, the wavefunction should satisfy momentum eigenvalue equation $$\hat p_x\psi(x)=-i\hbar\frac{d\psi}{dx}=p\psi(x)$$ It can be easily verified that both $\sin{kx}$ and $\cos{kx}$ are not eigenfunctions of $\hat p_x$. Hence, possible solutions are $$\psi(x)=Ae^{\pm ikx}$$ Out of these, for $k > 0$, $e^{ikx}$ represents particle moving from left to right ($+x$ direction) and $e^{-ikx}$ represents particle moving from right to left ($-x$ direction). Hence, possible solution is $$\psi(x)=Ae^{ikx}$$ Using normalization condition, we have $$|A|^2\int\limits_{0}^{L}dx=1$$ Hence, $A=\frac{1}{\sqrt{L}}$. \begin{align*} \psi(x)&=\frac{1}{\sqrt{L}}e^{ikx}\\ &=\frac{1}{\sqrt{L}}e^{i\sqrt{\frac{2mE}{\hbar^2}}x}\\ &=\frac{1}{\sqrt{L}}e^{i\frac{p}{\hbar}x} \end{align*}

Hence, answer is (D)

Problem set 50

1. Consider a energy level diagram shown below, which corresponds to the molecular nitrogen
   
   If the pump rate $R$ is $10^{20}$ atoms cm^-3s^-1 and the decay routes are as shown with $\tau_{12}=20\:ns$ and $\tau_1=1\:\mu s$, the equilibrium populations of states 2 and 1 are, respectively,
1. $10^{14}\: cm^{-3}$ and $2\times10^{12}\: cm^{-3}$
2. $2\times10^{12}\: cm^{-3}$ and $10^{14}\: cm^{-3}$
3. $2\times10^{12}\: cm^{-3}$ and $2\times10^{6}\: cm^{-3}$
4. zero, and $10^{20}\: cm^{-3}$
The differential rate equations can be written as $$\frac{dN_2}{dt}=R-\frac{N_2}{\tau_{12}}$$ $$\frac{dN_1}{dt}=\frac{N_2}{\tau_{12}}-\frac{N_1}{\tau_{1}}$$ For equlibrium, we have $$\frac{dN_2}{dt}=\frac{dN_1}{dt}=0$$ $$R-\frac{N_2}{\tau_{12}}=0$$ $$\frac{N_2}{\tau_{12}}-\frac{N_1}{\tau_{1}}=0$$ \begin{align*} N_2&=R\tau_{12}\\ &=20\times 10^{-9}\times 10^{20}\\ &=2\times10^{12}\:cm^{-3} \end{align*} \begin{align*} N_1&=\frac{\tau_{1}N_2}{\tau_{12}}\\ &=\frac{1\times10^{-6}\times2\times10^{12}}{20\times10^{-9}}\\ &=10^{14}\:cm^{-3} \end{align*}

Hence, answer is (B)

2. Using the Clausius-Clapeyron equation, the change in melting point of ice for 1 atmosphere rise in pressure is : (Given: The latent heat of fusion for water at $0^oC$ is $3.35\times10^5\:J/kg$, the volume of ice is $1.09070\:cc/g$ and the volume of water is $1.00013\: cc/gm$)
1. $-0.0075^oC$
2. $0.0075^oC$
3. $0.075^oC$
4. $-0.075^oC$
$$v_1=1.09070\: cc/gm$$ $$v_2=1.00013\: cc/gm$$ \begin{align*} L&=3.35\times10^5\:J/kg\\ &=80.069\: cal/gm\\ &=80.069\times4.2\times10^7 \:erg/gm \end{align*} $$T=0+273=273K$$ \begin{align*} dP&=1\: atm\\ &=76\times13.6\times980\:dyne/cm^2 \end{align*} Clausius-Clapeyron equation is $$\frac{dP}{dT}=\frac{L}{T(v_2-v_1)}$$ \begin{align*} dT&=\frac{T(v_2-v_1)dP}{L}\\ &={\textstyle\frac{273\times(1.00013-1.09070)\times76\times13.6\times980}{80.069\times4.2\times10^7}}\\ &=-0.0075\: ^oC \end{align*}

Hence, answer is (A)

3. A linear quadrupole is formed by joining two dipoles each of mmt $\vec P$ back-to-back. The electric potential and field at point P far away from the quadrupole is found to vary respectively with distance as :
1. $1/r^5$ and $1/r^4$
2. $1/r^2$ and $1/r^3$
3. $1/r^4$ and $1/r^3$
4. $1/r^3$ and $1/r^4$

Answer is (D)

4. Consider the elastic vibrations of a crystal with one atom in the primitive cell. If $m$ is mass of the atom, $a$ is the nearest neighbour distance and $c$ the force constant, the frequency of a lattice wave in terms of the wave vector $k$ is :
1. $\omega=\left(\frac{4c}{m}\right)^{\frac{1}{2}}\left|\sin{\frac{ka}{2}}\right|$
2. $\omega=\left(\frac{4c}{m}\right)^{\frac{1}{2}}\sin^2{\frac{ka}{2}}$
3. $\omega=\left(\frac{4c}{m}\right)^{\frac{1}{2}}\cos{\frac{ka}{2}}$
4. $\omega=\left(\frac{4c}{m}\right)^{\frac{1}{2}}\cos^2{\frac{ka}{2}}$
Force on $s^{th}$ plane is $$F_s=-c\left(u_s-u_{s-1}\right)-c\left(u_{s+1}-u_{s}\right)$$ Equation of motion is $$m\frac{d^2u_s}{dt^2}=-c\left(u_{s+1}+u_{s-1}-2u_{s}\right)$$ Wave solution with $u_s(t)=u_se^{-\omega t}$ and $u_s= u_0e^{ikas}$, we get $$\omega^2=\frac{2c}{m}(1-\cos{ka})$$ $$\omega=\left(\frac{4c}{m}\right)^{\frac{1}{2}}\left|\sin{\frac{ka}{2}}\right|$$

Hence, answer is (A)

5. Consider following particles: the proton $p$, the neutron $n$, the neutral pion $\pi^0$ and the delta resonance $\Delta^+$. When ordered of decreasing lifetime, the correct arrangement is as follows
1. $\pi^0, n, p, \Delta^+$
2. $ p,n, \Delta^+, \pi^0$
3. $ p,n,\pi^0, \Delta^+ $
4. $\Delta^+,n, \pi^0, p $
Mean-life:

Proton: stable,

Neutron: $885.7\pm0.8$ seconds,

$\pi^0$: $0.84\times10^{-16}$,

$\Delta^+$: $6\times10^{-24}$

Answer is (C)