Problem set 78

1. The experimentally measured spin $g$ factors of a proton and a neutron indicate that
1. both proton and neutron are elementary point particles
2. both proton and neutron are not elementary point particles
3. while proton is an elementary point particle, neutron is not
4. while neutron is an elementary point pat1icle, proton is not

both proton and neutron are not elementary point particles

Hence, answer is (B)

2. The tank circuit of a Hartley oscillator is shown in the figure. If $M$ is the mutual inductance between the inductors, the oscillation frequency is
   
1. $\frac{1}{2\pi\sqrt{(L_1+L_2+2M)C}}$
2. $\frac{1}{2\pi\sqrt{(L_1+L_2-2M)C}}$
3. $\frac{1}{2\pi\sqrt{(L_1+L_2+M)C}}$
4. $\frac{1}{2\pi\sqrt{(L_1+L_2-M)C}}$

Frequency of Hartley oscillator is given by $$f=\frac{1}{2\pi\sqrt{LC}}$$ where $L=L_1+L_2$ if the coils are assumed to be winded on different cores and $L_1+L_2+2M$ if they are winded on a single core.

Hence, answer is (A)

3. In the given digital logic circuit, $A$ and $B$ form the input. The output $Y$ is
   
1. $Y=\bar A$
2. $Y=A\bar B$
3. $Y=A\oplus B$
4. $Y=\bar B$

\begin{align*} Y&=\overline{\left(\bar A+B\right) \left( A+B\right)}\\ &=\overline{\left(\bar A+B\right)}+ \overline{\left( A+B\right)}\\ &=\overline{\left(\bar A\right)}\bar B+\bar A\bar B\\ &=A\bar B+\bar A\bar B\\ &=\left(A+\bar A\right)\bar B\\ &=\bar B\quad \text{since, }(A+\bar A=1) \end{align*}

Hence, answer is (D)

4. The largest analog output voltage from a 6-bit digital to analog converter (DAC) which produces 1.0 V output for a digital input of 010100, is
1. 1.6 V
2. 2.9 V
3. 3.15 V
4. 5.0 V

For a n-bit DAC $$V_{out}={\textstyle V_R\left(\frac{b_{n-1}}{2^1}+\frac{b_{n-2}}{2^2}+\cdots+\frac{b_{0}}{2^n}\right)}$$ For a 6-bit DAC $$V_{out}=\!\!{\textstyle V_R\!\!\left(\frac{b_{5}}{2^1}\!+\!\frac{b_{4}}{2^2}\!+\!\frac{b_{3}}{2^3}\!+\!\frac{b_{2}}{2^4}\!+\!\frac{b_{1}}{2^5}\!+\!\frac{b_{0}}{2^6}\right)}$$ For 1.0 V output of a digital input of 010100 $$1=\!{\textstyle V_R\!\!\left(\frac{0}{2^1}+\frac{1}{2^2}+\frac{0}{2^3}+\frac{1}{2^4}+\frac{0}{2^5}+\frac{0}{2^6}\right)}$$ $$V_R=\frac{16}{5}$$ $$V_{max}\!=\!{\textstyle \frac{16}{5}\!\left(\frac{1}{2^1}\!+\!\frac{1}{2^2}\!+\!\frac{1}{2^3}\!+\!\frac{1}{2^4}+\frac{1}{2^5}\!+\!\frac{1}{2^6}\right)}$$ $$V_{max}=3.15 V$$

Hence, answer is (C)

5. The low-pass active filter shown in the figure has a cut-off frequency of 2 kHz and a pass band gain of 1.5. The values of the resistors are
   
1. $R_1 = 10\: k\Omega$; $R_2 = 1.3 \Omega$
2. $R_1 = 30\: k\Omega$; $R_2 = 1.3 \Omega$
3. $R_1 = 10\: k\Omega$; $R_2 = 1.7 k\Omega$
4. $R_1 = 30\: k\Omega$; $R_2 = 1.7 k\Omega$

Using $$Gain=1+\frac{15k}{R_1}$$ $$R_1=30\:k\Omega$$ Using $$f_c=\frac{1}{2\pi RC}$$ $$R=\frac{1}{2\pi f_cC}=1.7\:k\Omega$$

Hence, answer is (D)

Problem set 77

1. The energy $E(\vec k)$ of electrons of wavevector $\vec k$ in a solid is given by $E(\vec k) = Ak^2 + Bk^4$, where $A$ and $B$care constants. The effective mass of the electron at $|\vec k| = k_0$ is
1. $Ak_0^2$
2. $\frac{\hbar^2}{2A}$
3. $\frac{\hbar^2}{2A+12Bk_0^2}$
4. $\frac{\hbar^2}{2A+12Bk_0^2}$

Effective mass is given by $$m*=\frac{\hbar^2}{\frac{d^2E}{dk^2}}$$ $$m*=\frac{\hbar^2}{2A+12Bk_0^2}$$

Hence, answer is (C)

2. Which one of the following statements is NOT correct about the Brillouin zones (BZ) of a square lattice with lattice constant a?
1. The first BZ is a square of side $2\pi/a$ in $k_x$-$k_y$ plane
2. The areas of the first BZ and third BZ are the same
3. The $k$-points are equidistant in $k_x$ as well as in $k_y$ directions
4. The area of the second BZ is twice that of the first BZ

The area of the second BZ is twice that of the first BZ is not correct

Hence, answer is (D)

3. In a crystal of $N$ primitive cells, each cell contains two monovalent atoms. The highest occupied energy band of the crystal is
1. one-fourth filled
2. one-third filled
3. half filled
4. completely filled

Each band can be occupied by $2N$ electrons, if there are $N$ primitive cells in a crystal. Hence, if each primitive cell contains 2 electrons, the highest occupied energy band of the crystal is completely filled.

Hence, answer is (D)

4. If the number density of a free electron gas changes from $10^{28}$ to $10^{26}$ electrons/$m^3$, the value of plasma frequency (in Hz) changes from $5.7\times 10^{15}$ to
1. $5.7\times 10^{13}$
2. $5.7\times 10^{14}$
3. $5.7\times 10^{16}$
4. $5.7\times 10^{17}$

Plasma frequency is given by $$\omega_p=2\pi f_p=\sqrt{\frac{ne^2}{m\epsilon_0}}$$ $$f_{p_1}\propto\sqrt{n_1}$$ $$f_{p_2}\propto\sqrt{n_2}$$ $$\frac{f_{p_2}}{f_{p_1}}=\sqrt{\frac{n_2}{n_1}}$$ $$f_{p_2}=f_{p_1}\sqrt{\frac{n_2}{n_1}}$$ $$f_{p_2}=5.7\times 10^{15}\times\sqrt{\frac{10^{26}}{10^{28}}}$$ $$f_{p_2}=5.7\times 10^{14}$$

Hence, answer is (B)

5. Which one of the following statements about superconductors is NOT true?
1. A type I superconductor is completely diamagnetic
2. A type II superconductor exhibits Meissner effect upto the second critical magnetic field ($H_{c_2}$)
3. A type II superconductor exhibits zero resistance upto the second critical magnetic field
4. Both type I and type II superconductors exhibit sharp fall in resistance at the superconducting transition temperature

In type II superconductor the transition from superconducting to normal state occurs after going through a broad "mixed state" region.

Hence, answer is (D)

Problem set 76

1. Which one of the following electronic transitions in neon is NOT responsible for LASER action in a helium-neon laser?
1. $6s\rightarrow5p$
2. $5s\rightarrow4p$
3. $5s\rightarrow3p$
4. $4s\rightarrow3p$

$6s\rightarrow5p$

Hence, answer is (A)

2. In the linear Stark effect, the application of an electric field
1. completely lifts the degeneracy of $n = 2$ level of hydrogen atom and splits $n = 2$ level into four levels
2. partially lifts the degeneracy of $n = 2$ level of hydrogen atom and splits $n = 2$ level into three levels
3. partially lifts the degeneracy of $n = 2$ level of hydrogen atom and splits $n = 2$ level into two levels
4. does not affect the $n= 2$ levels

Partially lifts the degeneracy of $n = 2$ level of hydrogen atom and splits $n = 2$ level into three levels

Hence, answer is (B)

3. In hyperfine interaction, there is coupling between the electron angular momentum $\vec J$ and nuclear angular momentum $\vec I$, forming resultant angular momentum $\vec F$. The selection rules for the corresponding quantum number $F$ in hyperfine transitions are
1. $\Delta F=\pm2$ only
2. $\Delta F=\pm1$ only
3. $\Delta F=0,\pm1$
4. $\Delta F=\pm1,\pm2$

$\Delta F=0,\pm1$

Hence, answer is (C)

4. A vibrational-electronic spectrum of homonuclear binary molecules, involving electronic ground state $\epsilon''$ and excited state $\epsilon'$, exhibits a continuum at $\bar\nu cm^{-1}$. If the total energy of the dissociated atoms in the excited state exceeds the total energy of the dissociated atoms in the ground state by $E_{ex} cm^{-1}$, the dissociation energy of the molecule in the ground state is
1. $(\bar\nu+E_{ex})/2$
2. $(\bar\nu-E_{ex})/2$
3. $(\bar\nu-E_{ex})$
4. $\sqrt{(\bar\nu^2-E_{ex}^2)}$

$D=(\bar\nu-E_{ex})$

Hence, answer is (C)

5. The NMR spectrum of ethanol $(CH_3CH_2OH)$ comprises of three bunches of spectral lines. The number of spectral lines in the bunch corresponding to $CH_2$ group is
1. 1
2. 2
3. 3
4. 4

In ethanol the intensities are in the ratio 3:2:1, because there are three $CH_3$ protons, two $CH_2$ protons and one $OH$ proton in a molecule. So the number of spectral lines in the bunch corresponding to $CH_2$ group is 2

Hence, answer is (B)

Problem set 75

1. The internal energy of $n$ moles of a gas is given by $E = \frac{3}{2}nRT- \frac{a}{V}$, where $V$ is the volume of the gas at temperature $T$ and $a$ is a positive constant. One mole of the gas in state $(T_1, V_1)$ is allowed to expand adiabatically into vacuum to a final state $(T_2, V_2)$. The temperature $T_2$ is
1. $T_1+Ra\left(\frac{1}{V_2}+\frac{1}{V_1}\right)$
2. $T_1-\frac{2}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$
3. $T_1+\frac{2}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$
4. $T_1-\frac{1}{3}Ra\left(\frac{1}{V_2}-\frac{1}{V_1}\right)$

According to first of thermodynamics $$dU=dQ+dW$$ For adiabatic process there is no heat transfer. Hence, $dQ=0$ $$dU=dW$$ For a free expansion of gas in vacuum, work done is zero, $dW=0$ $$dU=E_2-E_1=0$$ $$E_2=E_1$$ $$\frac{3}{2}nRT_2- \frac{a}{V_2}=\frac{3}{2}nRT_1- \frac{a}{V_1}$$ $$\frac{3}{2}nRT_2 =\frac{3}{2}nRT_1- \frac{a}{V_1}+\frac{a}{V_2}$$ $$T_2 =T_1+\frac{2a}{3nR}\left( \frac{1}{V_2}-\frac{1}{V_1}\right)$$ for one mole $n=1$ $$T_2 =T_1+\frac{2a}{3R}\left( \frac{1}{V_2}-\frac{1}{V_1}\right)$$

Hence, answer is (C)

2. A monatomic crystalline solid comprises of $N$ atoms, out of which $n$ atoms are in interstitial positions. If the available interstitial sites are $N'$, the number of possible microstates is
1. $\frac{(N'+n)!}{n!N!}$
2. $\frac{N!}{n!(N+n)!}\frac{N'!}{n!(N'+n)!}$
3. $\frac{N!}{n!(N'-n)!}$
4. $\frac{N!}{n!(N-n)!}\frac{N'!}{n!(N'-n)!}$

$\frac{N!}{n!(N-n)!}\frac{N'!}{n!(N'-n)!}$

Hence, answer is (D)

3. A system of $N$ localized, non-interacting spin $1/2$ ions of magnetic moment $\mu$ each is kept in an external magnetic field $H$. If the system is in equilibrium at temperature $T$, the Helmholtz free energy of the system is
1. $Nk_BT\ln{\left(\cosh{\frac{\mu H}{k_BT}}\right)}$
2. $-Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)}$
3. $Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)}$
4. $-Nk_BT\ln{\left(2\sinh{\frac{\mu H}{k_BT}}\right)}$

A spin $\frac{1}{2}$ has two possible orientations. (These are the two possible values of the projection of the spin on the z axis: $m_s=\pm1/2$.) Associated with each spin is a magnetic moment which has the two possible values $\pm\mu$. Hence, in a magnetic field the two states are of different energy ($E=-\vec\mu\cdot\vec H$). Hence, single particle partition function is given by \begin{align*} z&=\sum e^{-E/k_BT}\\ &=e^{-\mu H/k_BT}+e^{\mu H/k_BT}\\ &=2\cosh{\frac{\mu H}{k_BT}} \end{align*} Hence, N particle partition function is $$Z=z^N=\left(2\cosh{\frac{\mu H}{k_BT}}\right)^N$$ Helmholtz free energy is given by \begin{align*} F&=-k_BT\ln{Z}\\ &=-Nk_BT\ln{\left(2\cosh{\frac{\mu H}{k_BT}}\right)} \end{align*}

Hence, answer is (B)

4. The phase diagram of a free particle of mass $m$ and kinetic energy $E$, moving in a one-dimensional box with perfectly elastic walls at $x = 0$ and $x = L$, is given by


For free particle $E=\frac{p_x^2}{2m}$ $$p_x=\pm\sqrt{2mE}$$ Particle bounce forth and back between $x=0$ and $x=L$

Hence, answer is (A)

5. In the microwave spectrum of identical rigid diatomic molecules, the separation between the spectral lines is recorded to be $0. 7143\: cm^{-1}$. The moment of inertia of the molecule, in $kg\: m^2$, is
1. $2.3 \times 10^{-36}$
2. $2.3 \times 10^{-40}$
3. $7.8 \times 10^{-42}$
4. $7.8 \times 10^{-46}$

In rotational spectra separation between two spectral line is $2B=0. 7143\: cm^{-1}$ $$B=\frac{h}{8\pi^2cI}$$ When $B$ is in $cm^{-1}$, $c$ should be taken in $cm\:s^{-1}$. \begin{align*} I&=\frac{h}{4\pi^2c2B}\\ &=\frac{6.62607\times10^{-34}}{4\times3.14^2\times3\times10^{10}\times0.7143}\\ &=7.832\times10^{-46} \end{align*}

Hence, answer is (D)

Problem set 74

1. The wavefunction of a particle in a one-dimensional potential at time $t=0$ is $\psi(x,t=0)=\frac{1}{\sqrt{5}}\left[2\psi_0(x)-\psi_1(x)\right]$ where $\psi_0(x)$ and $\psi_1(x)$ are the ground and the first excited states of the particle with corresponding energies $E_0$ and $E_1$. The wavefunction of the particle at a time $t$ is
1. $\frac{1}{\sqrt{5}}e^{-i(E_0+E_1)t/2\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
2. $\frac{1}{\sqrt{5}}e^{-iE_0t/\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
3. $\frac{1}{\sqrt{5}}e^{-iE_1t/\hbar}\left[2\psi_0(x)-\psi_1(x)\right]$
4. ${\scriptstyle \frac{1}{\sqrt{5}}\left[2\psi_0(x)e^{-iE_0t/\hbar}-\psi_1(x)e^{-iE_1t/\hbar}\right]}$

$$\hat H\psi_0=E_0\psi_0\quad\hat H\psi_1=E_1\psi_1$$ Time evolution of quantum systems is given by Unitary Transformations, \begin{align*} \left|\psi(x,t)\right > &=\hat U \left|\psi(x,t=0)\right >\\ &=e^{-i\hat Ht/\hbar} \left|\psi(x,t=0)\right >\\ &={\scriptstyle e^{-i\hat Ht/\hbar} \frac{1}{\sqrt{5}}\left[2\psi_0(x)-\psi_1(x)\right]}\\ &={\scriptstyle \frac{1}{\sqrt{5}}\left[2e^{-i\hat Ht/\hbar}\psi_0(x)-e^{-i\hat Ht/\hbar}\psi_1(x)\right]}\\ \end{align*} Using property $f(\hat A)\psi_n=f(a_n)\psi_n$, we get $${\scriptstyle \left|\psi(x,t)\right > =\frac{1}{\sqrt{5}}\left[2\psi_0(x)e^{-iE_0t/\hbar}-\psi_1(x)e^{-iE_1t/\hbar}\right]}$$

Hence, answer is (D)

2. The commutator $[L_x, y]$, where $L_x$ is the $x$-component of the angular momentum operator and $y$ is the $y$-component of the position operator, is equal to
1. 0
2. $i\hbar x$
3. $i\hbar y$
4. $i\hbar z$

$$L_x=yp_z-zp_y$$ \begin{align*} [L_x, y]&=[yp_z-zp_y, y]\\ &=[yp_z, y]-[zp_y, y]\\ &={\scriptstyle y[p_z, y]+[y, y]p_z-z[p_y, y]-[z, y]p_y}\\ &=i\hbar z \end{align*}

Hence, answer is (D)

3. In hydrogenic states, the probability of finding an electron at $r = 0$ is
1. zero in state $\phi_{1s}(r)$
2. non-zero in state $\phi_{1s}(r)$
3. zero in state $\phi_{2s}(r)$
4. non-zero in state $\phi_{2p}(r)$

non-zero in state $\phi_{1s}(r)$

Hence, answer is (B)

4. Each of the two isolated vessels, $A$ and $B$ of fixed volumes, contains $N$ molecules of a perfect monatomic gas at a pressure $P$. The temperatures of $A$ and $B$ are $T_1$ and $T_2$, respectively. The two vessels are brought into thermal contact. At equilibrium, the change in entropy is
1. $\frac{3}{2}Nk_B\ln{\left[\frac{T_1^2+T_2^2}{4T_1T_2}\right]}$
2. $Nk_B\ln{\left(\frac{T_2}{T_1}\right)}$
3. $\frac{3}{2}Nk_B\ln{\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]}$
4. $2Nk_B$

Energy of $N$ monatomic gas molecules is given by $$E=\frac{3}{2}Nk_BT$$ $$E_A=\frac{3}{2}Nk_BT_1$$ $$E_B=\frac{3}{2}Nk_BT_2$$ $$C_A=C_B=C=\frac{3}{2}Nk_B$$ When two bodies are in thermal contact, final temperature is given by $$T_f=\frac{C_AT_1+C_BT_2}{C_A+C_B}$$ $$T_f=\frac{T_1+T_2}{2}$$ Change in entropy of body $A$ is \begin{align*} \Delta S_A&=\int_{T_1}^{T_f}\frac{dQ}{T}\\ &=\int_{T_1}^{T_f}\frac{CdT}{T}\\ &=\frac{3}{2}Nk_B\ln{\frac{T_f}{T_1}} \end{align*} $$\Delta S_B=\frac{3}{2}Nk_B\ln{\frac{T_f}{T_2}}$$ Total change in entropy is $$\Delta S=\Delta S_A+\Delta S_B$$ \begin{align*} \Delta S&=\frac{3}{2}Nk_B\left(\ln{\frac{T_f}{T_1}}+\ln{\frac{T_f}{T_2}} \right)\\ &=\frac{3}{2}Nk_B\ln{\left[\frac{T_f^2}{T_1T_2}\right]} \\ &=\frac{3}{2}Nk_B\ln{\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]} \\ \end{align*}

Hence, answer is (C)

5. The mean internal energy of a one-dimensional classical harmonic oscillator in equilibrium with a heat bath of temperature $T$ is
1. $\frac{1}{2}k_BT$
2. $k_BT$
3. $\frac{3}{2}k_BT$
4. $3k_BT$

Energy of harmonic oscillator is given by $$E=\frac{p_x^2}{2m}+\frac{1}{2}kx^2$$ According to equipartition theorem each quadratic degrees of freedom has energy $\frac{1}{2}k_BT$. Energy is quadratic in $p_x$ and also in $x$. Hence, $$E=\frac{1}{2}k_BT+\frac{1}{2}k_BT=k_BT$$

Hence, answer is (B)

Problem set 73

1. For the logic circuit given below, the decimal count sequence and the modulus of the circuit corresponding to A B C D are (here, J=K=1 locked)
   
1. $8\rightarrow4\rightarrow2\rightarrow1\rightarrow9\rightarrow5$ (mod 6)
2. $8\rightarrow4\rightarrow2\rightarrow9\rightarrow5\rightarrow3$ (mod 6)
3. $2\rightarrow5\rightarrow9\rightarrow1\rightarrow3$ (mod 5)
4. $8\rightarrow5\rightarrow1\rightarrow3\rightarrow7$ (mod 5)

In the T or "toggle" flip-flop both J=1 and k=1 and it changes its output on each clock edge, giving an output which is half the frequency of the signal to the T input.

Clock	A	B	C	D	Decimal
0	1	0	0	0	8
1	0	1	0	0	4
2	0	0	1	0	2
3	1	0	0	1	9
4	0	1	0	1	5
5	0	0	1	1	3
6	1	0	0	0	8

Modulus of a counter, is the number of states that the counter will sequence through before returning back to its original value.

Hence, answer is (B)

2. The low-energy electronic excitations in a two-dimensional sheet of graphene is given by $E(\vec k)=\hbar v k$, where $v$ is the velocity of the excitations. The density of states is proportional to
1. $E$
2. $E^{3/2}$
3. $E^{1/2}$
4. $E^{2}$

In two-dimensions area of k-space is $$ \Omega _{2,k}\propto k^2$$ Using $E(\vec k)=\hbar v k$ $$ \Omega _{2,k}\propto \frac{E^2}{\hbar^2 v^2}$$ $$D_n(E)=\frac{d\Omega _{n,k}}{dE}$$ $$D_2(E)=\frac{d\Omega _{2,k}}{dE}\propto E$$

Hence, answer is (A)

3. X-ray of wavelength $\lambda=a$ is reflected from $(111)$ plane of a simple cubic lattice. If the lattice constant is $a$, the corresponding Bragg angle (in radian) is
1. $\pi/6$
2. $\pi/4$
3. $\pi/3$
4. $\pi/8$

$$d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}}$$ $$d_{111}=\frac{a}{\sqrt{3}}$$ $$\sin\theta=\frac{\lambda}{2d_{111}}$$ $$\sin\theta=\frac{\sqrt{3}}{2}$$ $$\theta=\pi/3$$

Hence, answer is (C)

4. Let us approximate the nuclear potential in the shell model by a three dimensional isotropic harmonic oscillator. Since the lowest two energy levels have angular momenta $l=0$ and $l=1$ respectively, which of the following two nuclei have magic numbers of protons and neutrons?
1. $_2^4He$ and $_8^{16}O$
2. $_1^2D$ and $_4^{8}Be$
3. $_2^4He$ and $_4^{8}Be$
4. $_2^4He$ and $_6^{12}C$

The energy eigenvalues of the harmonic oscillator are $$E_n=(n+\frac{1}{2})\hbar\omega\quad n=1,2,3,\cdots$$ In case of harmonic oscillator for a given $n$ $$l=n-1,n-2,\cdots0$$ $$m_l=-l,-l+1,\dots l$$ $$m_s=\pm\frac{1}{2}$$ For $l=0$, $n=1$, $m_l=0$ and $m_s=\pm\frac{1}{2}$. Hence there are two states explaining the first magic number, 2. Two nucleons of a given type can occupy the lowest energy level. $_2^4He$ has the lowest energy level for protons completely filled with its two protons, and the lowest level for neutrons completely filled with its two neutrons. That makes $_2^4He$ the first doubly-magic nucleus.

For $l=1$, $n=2$, $m_l=-1, 0,1$ and $m_s=\pm\frac{1}{2}$. Therefore, $l=1$ at $n=2$ corresponds to 6 energy states. Combined with the two $l= 0$ states at energy level $n=1$, that gives a total of 8. The second magic number 8 has been explained! It requires 8 nucleons of a given type to fill the lowest two energy levels. It makes $_8^{16}O$ with 8 protons and 8 neutrons the second doubly-magic nucleus.

Hence, answer is (A)

5. The reaction $^2_1D+^2_1D\rightarrow ^4_2He+\pi^0$ cannot proceed via strong interactions because it violets the conservation of
1. angular momentum
2. electric charge
3. baryon number
4. isospin

A nucleus of deuteron contains one proton and one neutron (each with a baryon number of 1) and has a baryon number of 2 and charge +1. isospin=0, Owing to the generalized Pauli principle, it must have a symmetric spin state, i.e. $S= 1$, and a total angular momentum $J=1$. Helium nucleus is composed of two protons and two neutrons. Hence, Baryon number 4 and charge +2, isospin=0

	$^2_1D+^2_1D\rightarrow ^4_2He+\pi^0$
Q	$1+1\rightarrow 2+0$	Conserved
B	$2+2\rightarrow 4+0$	Conserved
Isospin	$0+0\rightarrow 0+1$	Not Conserved

Hence, answer is (D)

Problem set 72

1. In the circuit given below, the thermister has a resistance 3 $k\Omega$ at $25^o C$. Its resistance decreases by 150 $\Omega$ per $^oC$ upon heating. The output voltage of the circuit at $30^oC$ is
   
1. $-3.75$ V
2. $-2.25$ V
3. $2.25$ V
4. $3.75$ V

Resistance of thermister at $30^oC$ is $R_f=2.25$ \begin{align*} V_{out}&=-\frac{R_f}{R_i}V_{in}\\ &=-\frac{2.25}{1}(-1)\\ &=2.25\:V \end{align*}

Hence, answer is (C)

2. Of the following term symbols of the $np^2$ atomic configurations, $^1S_0$, $^3P_0$, $^3P_1$, $^3P_2$ and $^1D_2$, which is the ground state?
1. $^3P_0$
2. $^1S_0$
3. $^3P_2$
4. $^3P_1$

According to Hund’s rules

1. State with highest multiplicity has lowest energy
2. State with same multiplicity, the state with highest $L$ will have lowest energy
3. State with same multiplicity and $L$ value. The state with lowest $J$ has lowest energy (only if subshell is less than half filled)

From the given states $^1S_0$, $^3P_0$, $^3P_1$, $^3P_2$ and $^1D_2$, $^3P_0$ will have the lowest energy

Hence, answer is (A)

3. A diatomic molecule has vibrational states with energies $E_\nu = \hbar\omega \left(\nu+\frac{1}{2}\right)$ and rotational states with energies $E_j = Bj(j + 1)$, where $\nu$ and $j$ are non-negative integers. Consider the transitions in which both the initial and final states are restricted to $\nu\leq 1$ and $j\leq 2$ and subject to the selection rules $\Delta\nu = \pm 1$ and $\Delta j = \pm 1$. Then the largest allowed energy of transition is
1. $\hbar\omega-3B$
2. $\hbar\omega-B$
3. $\hbar\omega+4B$
4. $2\hbar\omega+B$

$$E=\hbar\omega \left(\nu+\frac{1}{2}\right)+Bj(j + 1)$$ where $j$ is lower quantum number For $\Delta\nu = \pm 1$ and $\Delta j = \pm 1$ $$\Delta E=E_i-E_f=\hbar\omega+2B(j + 1)$$ For the largest allowed energy of transition $j=1$ $$\Delta E=E_i-E_f=\hbar\omega+4B$$

Hence, answer is (C)

4. A He-Ne laser operates by using two energy levels of Ne separated by 2.26 eV. Under steady state conditions of optical pumping, the equivalent temperatures of the system at which the ratio of the number of atoms in the upper state to that in the lower state will be 1/20, is approximately (the Boltzman constant $k_B=8.6\times10^{-5}\:eV/K$
1. $10^{10}\:K$
2. $10^{8}\:K$
3. $10^{6}\:K$
4. $10^{4}\:K$

$$\frac{N_2}{N_1}=e^{-E/k_BT}$$ \begin{align*} T&=-\frac{E}{k_B}\frac{1}{\ln{\frac{N_2}{N_1}}}\\ &=-\frac{2.26}{8.6\times10^{-5}}\frac{1}{\ln{\frac{1}{20}}}\\ &\approx 10^{4} \end{align*}

Hence, answer is (D)

5. The critical magnetic fields of a superconductor at temperatures 4K and 8K are 11mA/m and 5.5 mA/m respectively. The transition temperature is approximately
1. 8.4 K
2. 10.6 K
3. 12.9 K
4. 15.0K

$$H_{c_T}=H_{c_0}\left(1-\left(\frac{T}{T_c}\right)^2\right)$$ $$H_{c_{T_1}}=H_{c_0}\left(1-\left(\frac{T_1}{T_c}\right)^2\right)$$ $$H_{c_{T_2}}=H_{c_0}\left(1-\left(\frac{T_2}{T_c}\right)^2\right)$$ $$\frac{H_{c_{T_1}}}{H_{c_{T_2}}}=\frac{T_c^2-T_1^2}{T_c^2-T_2^2}$$ $$\frac{T_c^2-16}{T_c^2-64}=\frac{11}{5.5}$$ $$T_c=10.6 \:K$$

Hence, answer is (B)

Problem set 71

1. A rod of length $L$ carries a total charge $Q$ distributed uniformly. If this is observed in a frame moving with a speed $v$ along the rod, the change per unit length (as measured by the moving observer) is
1. $\frac{Q}{L}\left(1-\frac{v^2}{c^2}\right)$
2. $\frac{Q}{L}\sqrt{1-\frac{v^2}{c^2}}$
3. $\frac{Q}{L\sqrt{1-\frac{v^2}{c^2}}}$
4. $\frac{Q}{L\left(1-\frac{v^2}{c^2}\right)}$

Length contraction is given by $$L'=L \sqrt{1-\frac{v^2}{c^2}}$$ Hence, charge per unit length is given by $$\frac{Q}{L\sqrt{1-\frac{v^2}{c^2}}}$$

Hence, answer is (C)

2. The electric and magnetic fields in the charge free region $z > 0$ are given by $$\vec E(\vec r,t)=E_0e^{-k_1z}\cos{(k_2x-\omega t)}\hat j$$ $${\scriptstyle\vec B(\vec r,t)=\frac{E_0}{\omega}e^{-k_1z}\left[k_1\!\sin{(k_2x-\omega t)}\hat i+k_2\!\cos{(k_2x-\omega t)}\hat k\right]}$$ where $\omega$, $k_1$, and $k_2$ are positive constants. The average energy flow in $x$-direction is
1. $\frac{E_0^2k_2}{2\mu_0\omega}e^{-2k_1z}$
2. $\frac{E_0^2k_2}{\mu_0\omega}e^{-2k_1z}$
3. $\frac{E_0^2k_1}{2\mu_0\omega}e^{-2k_1z}$
4. $\frac{1}{2}c\epsilon_0E_0^2e^{-2k_1z}$

According to Poynting theorem the energy flux is given by $$\vec S=\frac{1}{\mu_0}\vec E\times \vec B$$ Hence, \begin{align*} S_x\hat i&=\frac{1}{\mu_0}\left(E_0e^{-k_1z}\cos{(k_2x-\omega t)}\right)\hat j\times\\ &~~~~\left(\frac{E_0}{\omega}e^{-k_1z}k_2\cos{(k_2x-\omega t)} \right)\hat k\\ &=\frac{E_0^2k_2}{\mu_0\omega}e^{-2k_1z}\cos^2{(k_2x-\omega t)}\hat i\\ \end{align*} $$\left < S_x \right > =\frac{E_0^2k_2}{2\mu_0\omega}e^{-2k_1z}$$

Hence, answer is (A)

3. A uniform magnetic field in the positive $z$-direction passes through a circular wire loop of radius 1 cm and resistance 1 $\omega$ lying in $xy$-plane. The field strength is reduced from 10 tesla to 9 tesla in 1 s. The charge transferred across any point in the wire is approximately
1. $3.1\times10^{-4}$ coulomb
2. $3.4\times10^{-4}$ coulomb
3. $4.2\times10^{-4}$ coulomb
4. $5.2\times10^{-4}$ coulomb

Magnetic flux through circular wire is \begin{align*} \phi&=\int B\:dA\\ &=B\pi r^2\\ &=B\times3.14\times1\times10^{-4}\\ &=3.14\times10^{-4}B\\ \end{align*} Change in magnetic flux per unit time is \begin{align*} \frac{d\phi}{dt}&=3.14\times10^{-4}(B_2-B_1)\\ &=3.14\times10^{-4}(10-9)\\ &=3.14\times10^{-4} \end{align*} Faraday's law of induction can now be used to determine the induced emf: $$\epsilon=-\frac{d\phi}{dt}=-3.14\times10^{-4}\:volts$$ Current $$I=\frac{\epsilon}{R}=-3.14\times10^{-4}\:Amp$$ Hence, charge transferred in one second at point is $3.14\times10^{-4}$.

Hence, answer is (A)

4. A particle of mass $m$ is in a potential $V=\frac{1}{2}m\omega^2 x^2$, where $\omega$ is a constant. Let $\hat a=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{i\hat p}{m\omega}\right)$. In the Heisenberg picture $\frac{d\hat a}{dt}$ is given by
1. $\omega\hat a$
2. $-i\omega\hat a$
3. $\omega\hat a^\dagger$
4. $i\omega\hat a^\dagger$

In Heisenberg picture the time evolution of operator is given by $$\frac{d\hat A_H(t)}{dt}=\frac{i}{\hbar}\left[\hat H,\hat A_H(t)\right]+\frac{\partial\hat A_H(t)}{\partial t}$$ As $\hat A_H(t)$ does not explicitly depend on time $$\frac{d\hat A_H(t)}{dt}=\frac{i}{\hbar}\left[\hat H,\hat A_H(t)\right]$$ For harmonic oscillator $$\hat H=\frac{\hat p^2}{2m}+\frac{1}{2}m\omega^2 x^2$$ \begin{align*} \frac{d\hat p}{dt}&=\frac{i}{\hbar}\left[\hat H,\hat p\right]\\ &=-m\omega^2\hat x \end{align*} \begin{align*} \frac{d\hat x}{dt}&=\frac{i}{\hbar}\left[\hat H,\hat x\right]\\ &=\frac{\hat p}{m} \end{align*} \begin{align*} \frac{d\hat a}{dt}&=\sqrt{\frac{m\omega}{2\hbar}}\left(\frac{d\hat x}{dt}+\frac{i}{m\omega}\frac{d\hat p}{dt}\right)\\ &=\sqrt{\frac{m\omega}{2\hbar}}\left(\frac{\hat p}{m}-\frac{i}{m\omega}m\omega^2\hat x\right)\\ &=-i\omega\hat a \end{align*}

Hence, answer is (B)

5. Two different sets of orthogonal basis vectors $\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right\}$ and $\left\{\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix},\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\right\}$ are given for a two-dimensional real vector space. The matrix representation of a linear operator $\hat A$ in these bases are related by a unitary transformation. The unitary matrix may be chosen to be
1. $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$
2. $\begin{pmatrix}0&1\\1&0\end{pmatrix}$
3. $\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$
4. $\frac{1}{\sqrt{2}}\begin{pmatrix}1&0\\1&1\end{pmatrix}$

$$U_{ij}=< \psi_{new_i}|\psi_{old_j} >$$ \begin{align*} U_{11}&=< \psi_{new_1}|\psi_{old_1} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*} \begin{align*} U_{12}&=< \psi_{new_1}|\psi_{old_2} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*} \begin{align*} U_{21}&=< \psi_{new_2}|\psi_{old_1} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*} \begin{align*} U_{22}&=< \psi_{new_2}|\psi_{old_2} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\\ &=-\frac{1}{\sqrt{2}} \end{align*} $$U=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$

Hence, answer is (C)

Problem set 70

1. Consider the differential equation $\frac{dy}{dx}=x^2-y$ with initial condition $y=2$ at $x=0$. Let $y_{(1)}$ and $y_{(1/2)}$ be the solutions at $x=1$ obtained using Euler's forward algorithm with step size 1 and $\frac{1}{2}$ respectively. The value of $\left(y_{(1)}-y_{(1/2)}\right)/y_{(1/2)}$ is
1. $-1/2$
2. $-1$
3. $1/2$
4. 1

Euler's forward formula is $$y_{n+1}=y_n+hf(x_n,y_n)$$ where, $f(x,y)=x^2-y$. \begin{align*} y_{1}&=y_0+f(x_0,y_0)\\ &=2+f(0,2)\\ &=0 \end{align*} \begin{align*} y_{1/2}&=y_0+\frac{1}{2}f(x_0,y_0)\\ &=2+\frac{1}{2}f(0,2)\\ &=1 \end{align*} $$\left(y_{(1)}-y_{(1/2)}\right)/y_{(1/2)}=-1$$

Hence, answer is (B)

2. Let $f(x,t)$ be a solution of the wave equation $\frac{\partial^2f}{\partial t^2}=v^2\frac{\partial^2f}{\partial x^2}$ in 1-dimension. If at $t=0$, $f(x,0)=e^{-x^2}$ and $\frac{\partial f}{\partial t}(x,0)=0$ for all $x$, then $f(x,t)$ for all future times $t > 0$ is described by
1. $e^{-(x^2-v^2t^2)}$
2. $e^{-(x-vt)^2}$
3. $\frac{1}{4}e^{-(x-vt)^2}+\frac{3}{4}e^{-(x+vt)^2}$
4. $\frac{1}{2}\left[e^{-(x-vt)^2}+e^{-(x+vt)^2}\right]$

According to d’Alambert’s formula solution of wave equation is given by \begin{align*} f(x,t)&=\frac{f(x-vt,0)+f(x+vt,0)}{2}\\ &+\frac{1}{2v}\int\limits_{x-vt}^{x+vt}\frac{\partial f(x,0)}{\partial t}dx\\ &=\frac{\left[e^{-(x-vt)^2}+e^{-(x+vt)^2}\right]}{2}\\ &+\frac{1}{2v}\int\limits_{x-vt}^{x+vt}0 dx\\ &=\frac{\left[e^{-(x-vt)^2}+e^{-(x+vt)^2}\right]}{2} \end{align*}

Hence, answer is (D)

3. Let $q$ and $p$ be canonical coordinate and momentum of a dynamical system. Which of the following transformation is canonical?

   A: $Q_1=\frac{1}{\sqrt{2}}q^2$ and $P_1=\frac{1}{\sqrt{2}}p^2$

   B: $Q_2=\frac{1}{\sqrt{2}}(p+q)$ and $P_2=\frac{1}{\sqrt{2}}(p-q)$

1. neither A nor B
2. both A and B
3. only A
4. only B

A transformation is said be canonical if it satisfies Poisson Bracket $\left\{Q,P\right\}_{q,p}=1$ i.e. $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=1$$ Clearly, $$\frac{\partial Q_2}{\partial q}\frac{\partial P_2}{\partial p}-\frac{\partial Q_2}{\partial p}\frac{\partial P_2}{\partial q}=1$$

Hence, answer is (D)

4. The differential cross-section for scattering by a target is given by $\frac{d\sigma}{d\Omega}(\theta,\phi)=a^2+b^2\cos^2\theta$. If $N$ is the flux of the incoming particles, the number of particles scattered per unit time is
1. $\frac{4\pi}{3}N(a^2+b^2)$
2. $4\pi N(a^2+\frac{1}{6}b^2)$
3. $4\pi N(\frac{1}{2}a^2+\frac{1}{3}b^2)$
4. $4\pi N(a^2+\frac{1}{3}b^2)$

$$\frac{d\sigma}{d\Omega}(\theta,\phi)=a^2+b^2\cos^2\theta$$ \begin{align*} \sigma&=\int\frac{d\sigma}{d\Omega}d\Omega\\ &=\int\limits_0^{2\pi}\int\limits_0^{\pi}\frac{d\sigma}{d\Omega}\sin\theta d\theta d\phi\\ &=\int\limits_0^{2\pi}d\phi\int\limits_0^{\pi}\left(a^2+b^2\cos^2\theta\right)\sin\theta d\theta \\ &=\!2\pi\!\!\left[\!a^2\!\!\!\int\limits_0^{\pi}\!\!\sin\theta d\theta+b^2\!\!\!\int\limits_0^{\pi}\!\!\cos^2\!\theta\sin\theta d\theta\right] \\ &=2\pi\left[2a^2+b^2\int\limits_0^{\pi}\cos^2\theta\sin\theta d\theta\right] \\ &=2\pi\left[2a^2+b^2\int\limits_{-1}^{1}x^2 dx\right] \\ &=2\pi\left[2a^2+\frac{2}{3}b^2\right] \\ \sigma&=4\pi\left[a^2+\frac{1}{3}b^2\right] \\ \end{align*} Hence, number of particles scattered $=4\pi N\left[a^2+\frac{1}{3}b^2\right]$

Hence, answer is (D)

5. Consider a rectangular wave guide with transverse dimensions 2 m $\times$ 1 m driven with an angular frequency $\omega=10^9\: rad/s$. Which transverse electric (TE) modes will propagate in this wave guide?
1. $TE_{10}$, $TE_{01}$, and $TE_{20}$
2. $TE_{10}$, $TE_{11}$, and $TE_{20}$
3. $TE_{01}$, $TE_{10}$, and $TE_{11}$
4. $TE_{01}$, $TE_{10}$, and $TE_{22}$

Cut off frequency for rectangular waveguide is given by $$f_{c_{mn}}=\frac{v}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}$$ where, $v=\frac{1}{\sqrt{\epsilon_0\mu_0}}\frac{1}{\sqrt{\epsilon_r}}=\frac{c}{\sqrt{\epsilon_r}}$ is velocity propagation in dielectric.

The cutoff frequency for $TE_{mn}$ mode is the operating frequency below which attenuation occurs and above which propagation takes place.

For air filled wave guide $v=c$ $$f_{c_{mn}}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}$$ Now, $\omega=10^9\: rad/s$, hence, $f=\frac{\omega}{2\pi}=1.6\times10^8\:1/sec$ \begin{align*} f_{c_{22}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{2}{2}\right)^2+\left(\frac{2}{1}\right)^2}\\ &=3.35\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{20}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{2}{2}\right)^2+\left(\frac{0}{1}\right)^2}\\ &=1.5\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{11}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{1}\right)^2}\\ &=1.68\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{10}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{0}{1}\right)^2}\\ &=0.75\times10^8\:1/sec \end{align*} \begin{align*} f_{c_{01}}&=\frac{3\times10^8}{2}\sqrt{\left(\frac{0}{2}\right)^2+\left(\frac{1}{1}\right)^2}\\ &=1.5\times10^8\:1/sec \end{align*} As, $f < f_{c_{22}}$ and $f < f_{c_{11}}$, these modes will attenuate and as $f < f_{c_{22}}$, $f > f_{c_{10}}$, $f > f_{c_{01}}$, these modes will propagate.

Hence, answer is (A)

Problem set 69

1. A particle moves in two dimensions on the ellipse $x^2+4y^2=8$. At a particular instant it is at the point $(x,y)=(2,1)$ and the $x$-component of its velocity is 6 (in suitable units). Then the $y$-component of its velocity is
1. -3
2. -2
3. 1
4. 4

$$x^2+4y^2=8$$ differentiating with respect to $t$ $$2x\frac{dx}{dt}+8y\frac{dy}{dt}=0$$ At point $(x,y)=(2,1)$, $\frac{dx}{dt}=6$. Hence, $$2\times 2\times 6+8\times 1\times\frac{dy}{dt}=0$$ $$\frac{dy}{dt}=-3$$

Hence, answer is (A)

2. a particle of mass $m$ moves in the one-dimensional potential $V(x)=\frac{\alpha}{3}x^3+\frac{\beta}{4}x^4$ where $\alpha,\beta > 0$. One of the equilibrium points is $x=0$. The angular frequency of small oscillations about the other equilibrium point is
1. $\frac{2\alpha}{\sqrt{3m\beta}}$
2. $\frac{\alpha}{\sqrt{m\beta}}$
3. $\frac{\alpha}{\sqrt{12m\beta}}$
4. $\frac{\alpha}{\sqrt{24m\beta}}$

$$V(x)=\frac{\alpha}{3}x^3+\frac{\beta}{4}x^4$$ $$\frac{dV}{dx}=\alpha x^2+\beta x^3$$ At equilibrium points $\frac{dV}{dx}=0$, which gives $$\alpha x^2+\beta x^3=0$$ $$\Rightarrow x=0, x=-\frac{\alpha}{\beta}$$ For small oscillations, system performs SHM about equilibrium points. Hence, $$F=-kx\Rightarrow -\frac{dV}{dx}=-kx$$ $$\Rightarrow k=\frac{d^2V}{dx^2}$$ $$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{\frac{d^2V}{dx^2}}{m}}$$ $$\frac{d^2V}{dx^2}=2\alpha x+3\beta x^2$$ \begin{align*} \left.\frac{d^2V}{dx^2}\right|_{x=-\frac{\alpha}{\beta}}&=-\frac{2\alpha^2}{\beta}+\frac{3\alpha^2}{\beta}\\ &=\frac{\alpha^2}{\beta} \end{align*} $$\omega=\frac{\alpha}{\sqrt{m\beta}}$$

Hence, answer is (B)

3. A particle of unit mass moves in the $xy$-plane in such a way that $\dot x(t)=y(t)$ and $\dot y(t)=-x(t)$. We can conclude that it is in a conservative force-field which can be derived from the potential
1. $\frac{1}{2}(x^2+y^2)$
2. $\frac{1}{2}(x^2-y^2)$
3. $x+y$
4. $x-y$

For a conservative force field, we have, $$m\ddot{\vec r}=-\vec\nabla V$$ For $m=1$ $$\ddot{\vec r}=-\vec\nabla V$$ $$\ddot x\hat i+\ddot y\hat j=-\left(\frac{\partial V}{\partial x}\hat i+\frac{\partial V}{\partial y}\hat j\right)$$ Using $\dot x(t)=y(t)$ and $\dot y(t)=-x(t)$ in above equation we get $$ x\hat i+ y\hat j=\frac{\partial V}{\partial x}\hat i+\frac{\partial V}{\partial y}\hat j$$ Clearly, $V=\frac{1}{2}(x^2+y^2)$ satisfies above condition.

Hence, answer is (A)

4. The concentration of electrons, $n$, and holes, $p$, for an intrinsic semiconductor at a temperature $T$ can be expressed as $n=p=AT^{3/2}\exp{\left(-\frac{E_g}{2k_BT}\right)}$, where $E_g$ is the band gap and $A$ is a constant. If the mobility of both types of carriers is proportional to $T^{-3/2}$, then the log of the conductivity is a linear function of $T^{-1}$, with slope
1. $E_g/(2k_B)$
2. $E_g/k_B$
3. $-E_g/(2k_B)$
4. $-E_g/k_B$

Conductivity is given by $$\sigma=ne\mu$$ $$\sigma=AT^{3/2}\exp{\left(-\frac{E_g}{2k_BT}\right)}e\mu$$ But $\mu\propto T^{-3/2}$ $$\sigma=Be\exp{\left(-\frac{E_g}{2k_BT}\right)}$$ $$\ln\sigma=\ln{Be}-\frac{E_g}{2k_BT}$$ slope$=-\frac{E_g}{2k_B}$

Hence, answer is (C)

5. The rank-2 tensor $x_ix_j$, where $x_i$ are the Cartesian coordinates of the position vector in three dimensions, has 6 independent elements. Under rotation, these 6 elements decompose into irreducible sets (that is, the elements of each set transform only into linear combinations of elements in that set) containing
1. 4 and 2 elements
2. 5 and 1 elements
3. 3, 2 and 1 elements
4. 4, 1 and 1 elements

The rank-2 tensor $T_{ij}=x_ix_j$ can be decomposed into into irreducible representations in the following way $$T_{ij}=T_{ij}^{(0)}+T_{ij}^{(1)}+T_{ij}^{(2)}$$ where,

$T_{ij}^{(0)}$ is a tensor of rank 0 and has 1 independent component only.

$T_{ij}^{(1)}$ is a tensor of rank 1 and has 3 independent components.

$T_{ij}^{(2)}$ is a tensor of rank 2 and has 5 independent components.

Hence, answer is (B)

Problem set 68

1. In a counting experiment to determine the statistics obeyed by the $\beta$ particles emitted by a radioactive substance, the number of $\beta$ particles counted in 50 seconds time interval was repeatedly measured 100 times. The statistical ensemble in this case consists of the following number of members:
1. 50
2. 100
3. 5000
4. 2

As the reading is repeated 100 times, there will be 100 members in the ensembles.

Hence, answer is (B)

2. NaI(Tl) scintillation detector is used only for the detection of gamma radiation because
1. it has large scattering cross-section
2. it has small Comption scattering
3. it has small absorption cross-section
4. it has large absorption cross-section

It has large absorption cross-section

Hence, answer is (D)

3. Consider the elastic vibrations of a crystal with two atoms per primitive cell having masses $m$ and $M$ $(M > m)$. Using dispersion relation between the angular frequency $(\omega)$ and wave vector $(K)$, find the frequency for acoustic branch at $K=0$:
1. $\omega=Ka\left(\frac{2C}{m+M}\right)^{1/2}$
2. $\omega=2Ka\left(\frac{2C}{m+M}\right)^{1/2}$
3. $\omega=\frac{1}{2}Ka\left(\frac{2C}{m+M}\right)^{1/2}$
4. $\omega=\frac{1}{2}Ka\left(\frac{2C}{M}\right)^{1/2}$

The dispersion relation is given by $${\scriptstyle m M \omega^4-2C (m +M)\omega^2+2C^2(1−\cos{Ka})=0}$$ For $K\rightarrow0$, $\cos{Ka}=1-\frac{1}{2}K^2a^2$ $${\scriptstyle m M \omega^4-2C (m +M)\omega^2+C^2K^2a^2=0}$$ $${\textstyle \omega^2=2C\left(\frac{1}{m}+\frac{1}{M}\right)}\quad\text{Optical branch}$$ $$\omega^2=\frac{C^2K^2a^2}{2(m+M)}\quad\text{Acoustic branch}$$ $$\omega=\frac{1}{2}Ka\left(\frac{2C}{m+M}\right)^{1/2}$$

Hence, answer is (C)

4. In nuclear direct reactions, time of interaction is of the order of :
1. $10^{-10}$ sec
2. $10^{-16}$ sec
3. $10^{-22}$ sec
4. $10^{-30}$ sec

In nuclear direct interaction time must be very short :$10^{-22}$ sec

In compound nuclear interaction time is large :$10^{-18}$ to $10^{-16}$ sec

Hence, answer is (C)

5. The following decay states a conservation law that forbids it because: $$n\rightarrow p+e^-$$
1. conservation of angular momentum and conservation of Lepton numbers are both violated
2. conservation of baryon number and conservation of Lepton number are both violated
3. conservation of energy is violated
4. conservation of electric charge is violated

$$n\rightarrow p+e^-$$

1. There is conversation of electric charge as protron has charge $+e$, electron has charge $-e$ and neutron is charge-less.
2. baryon number is conserved since baryon number of proton is $+1$, of neutron is $+1$ and of electron is $0$
3. conservation of Lepton number is violated since Lepton number of proton is $0$, of neutron is $0$ and of electron is $1$
4. conservation of angular momentum is also violated since, when the decay of the neutron into a proton and an electron was observed, it did not fit the pattern of two-particle decay. That is, the electron emitted does not have a definite energy as is required by conservation of energy and momentum for a two-body decay.

Hence, answer is (A)

Problem set 67

1. The far infrared rotational absorption spectrum of a diatomic molecule shows equidistant lines with a spacing 20 cm$^{-1}$. The position of the first Stokes line in the rotational Raman spectrum of this molecule is:
1. 20 cm$^{-1}$
2. 40 cm$^{-1}$
3. 60 cm$^{-1}$
4. 120 cm$^{-1}$

The energies of the rotational levels are given by $$E_J=J(J+1)\frac{\hbar^2}{2I}$$ The transition energies for absorption of radiation are given by \begin{align*} h\nu&=E_f-E_i\\ &=J_f(J_f+1)\frac{\hbar^2}{2I}-J_i(J_i+1)\frac{\hbar^2}{2I} \end{align*} But selection rule for transition is $\Delta J=\pm1$. Hence, $J_f=J_i+1$. $$\nu=2(J_i+1)\frac{\hbar^2}{2Ih}$$ In terms of wavenumber $$\bar\nu=2(J_i+1)\frac{\hbar^2}{2Ihc}$$ $$\bar\nu=2B(J_i+1)$$ where, $B=\frac{\hbar^2}{2Ihc}$

The lowest energy transition is between $J_i = 0$ and $J_f = 1$ so the first line in the spectrum appears at a frequency of $2B$. The next transition is from $J_i = 1$ to $J_f = 2$ so the second line appears at $4B$. The spacing of these two lines is $2B$. In fact the spacing of all the lines is $2B$ according to this equation. $$2B=20\: cm^{-1}\Rightarrow B=10$$ Stokes lines are observed at $$\bar\nu=\bar\nu_0-B(4J+6)\:cm^{-1}$$ Hence, there is a gap of $6B$ between $\bar\nu_0$ and $1^{st}$ Stokes line. $$\bar\nu_0-\bar\nu=6B=60\:cm^{-1}$$

Hence, answer is (C)

2. Consider an infinite line of ions of alternating sign. If a distance between the adjacent ions is $R$, the Madelung constant for this chain of ions is :
1. $4\log4$
2. $4\log2$
3. $2\log2$
4. $2\log4$

Let us pick up a positive ion for reference. This ion has two negative ions as its neighbours on either side at a distance $R$, two positive ions as its next neighbours on either side at a distance $2R$, and so on. Hence, total energy due to all ions is \begin{align*} &V=-\frac{2e^2}{R}+\frac{2e^2}{2R}-\frac{2e^2}{3R}+\frac{2e^2}{4R}\cdots\\ &=-\frac{2e^2}{R}\left[2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)\!\right]\\ &=-\frac{2e^2}{R}\left[2\log2\right]\\ \end{align*}

Hence, answer is (C)

3. The trivalent gadolinimum ion has seven electrons in its outer orbital. The Lande $g$ factor for this ion is:
1. 1
2. $\frac{3}{2}$
3. 2
4. $\frac{5}{2}$

The trivalent gadolinimum has electron configuration $[Xe]4f^7$. Hence, there are 7 electrons in $f$ orbital. For $f^7$ configuration we have

	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$
$m_l=$	+3	+2	+1	0	-1	-2	-3

Hence, \begin{align*} L&={\scriptstyle(+3)+(+2)+(+1)+(0)+(-1)+(-2)+(-3)}\\ &=0 \end{align*} i.e. $S$ state \begin{align*} S&={\scriptstyle\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}}\\ &=7/2 \end{align*} $$2S+1=8$$ $$J=L+S=0+7/2=7/2$$ Term symbol is $^8S_{7/2}$ \begin{align*} g_J\!&=\!1\!+\!\frac{J(J\!+\!1)\!+\!S(S\!+\!1)\!-\!L(L\!+\!1)}{2J(J\!+\!1)}\\ &=1+\frac{\frac{7}{2}(\frac{7}{2}+1)+\frac{7}{2}(\frac{7}{2}+1)-0}{2\frac{7}{2}(\frac{7}{2}+1)}\\ &=2 \end{align*}

Hence, answer is (C)

4. An n-type semiconductor has an electron concentration of $3\times10^{20}/m^3$. If the electron drift velocity is $100\:m/s$ in an electric field of $200\:V/m$, the conductivity of this material (in units of $\Omega^{-1}m^{-1}$) is :
1. 24
2. 36
3. 48
4. 96

Drift velocity $\bar v=-\mu_e\bar E$

Charge density $\rho=-nq$, where $n$ is electron concentration

Current density $\bar J=\rho\bar v=-nq\bar v=nq\mu_e \bar E$

conductivity $\sigma=nq\mu_e$ \begin{align*} \sigma&=nq\mu_e\\ &=\frac{nqv}{E}\\ &=\frac{3\times10^{20}\times1.6\times10^{-19}\times100}{200}\\ &=24 \end{align*}

Hence, answer is (A)

5. If the temperature of a black body enclosure is tripled, the number of photons will increase by a factor of :
1. 2
2. 9
3. 8
4. 27

Number of photons enclosed in black body are given by $$N=\left(\frac{16\pi k^3\zeta(3)}{c^3h^3}\right)VT^3$$ where, $V$ is volume and $T$ is temperature of black body and $\zeta(3)$ is Riemann zeta function. Hence, $$N\propto VT^3$$

Hence, answer is (D)

Problem set 66

1. Consider a configuration of a system of 10 distinguishable particles in which there are 3 particles in state 1, 3 particles in state 2 and 4 particles in state 3. The total number of microstates is :
1. 4200
2. 864
3. 102060
4. 360

Number of ways of arranging $N$ distinguishable particles into groups of $N_j$ each, such that $N_1$ in state 1, $N_2$ in state 2, etc. and $N_j$ in the $j^{th}$ state are given by $$W(N)=\frac{N!}{N_1!N_2!\dots N_j!}$$ $$W(10)=\frac{10!}{3!\times 3!\times 4!}=4200$$

Hence, answer is (A)

2. In the density matrix representation, the condition for a pure state is:
1. $\hat\rho^2=\hat\rho$
2. $\hat\rho^2=\hat I$
3. $T_r\hat\rho^2=T_r\hat\rho$
4. $T_r\hat\rho=1$

The Density matrix for the pure state $|\psi>$ is given by $$\rho=|\psi><\psi|$$ This density matrix has the following properties:

1. $\hat\rho^2=\hat\rho$ projector
2. $\hat\rho^\dagger=\hat\rho$ hermiticity
3. $T_r\hat\rho=1$ normalization
4. $\hat\rho\geq 0$ positivity

Hence, answer is (A)

3. It is required to operate a G.M. counter with a maximum radial field $10^7$ V/m. The applied voltage required if the radii of the wire and tube are 0.002 cm and 1 cm respectively.
1. $10^7$ volts
2. $1242\times 10^7$ volts
3. $1242$ volts
4. $12$ volts

In a cylindrical geometry with the anode at the center, the electric field at radius $r$ from the anode is given by $$E(r)=\frac{V}{r\ln{\left(\frac{b}{a}\right)}}$$ where $V =$ the applied voltage, $a =$ anode wire radius, $b =$ cathode tube inner radius. $$V=E(r)\times r\times\ln{\left(\frac{b}{a}\right)}$$ $E(r)$ is maximum on the anode surface, hence, $r=0.002 \:cm=2\times10^{-5}m$ $$V=10^{7}\times 2\times10^{-5}\times\ln{\left(\frac{1}{0.002}\right)}$$ $$V=1242\:volts$$

Hence, answer is (C)

4. Analysis of a X-ray diffraction pattern of a material crystallizing in a fcc-type structure gives a value of lattice constant as 'a'. The nearest neighbour distance in the material is:
1. $a$
2. $\sqrt{3}\frac{a}{2}$
3. $\frac{a}{\sqrt{2}}$
4. $\frac{a}{2}$

In FCC-lattice nearest neighbour distance is $\frac{a}{\sqrt{2}}$

Hence, answer is (C)

5. Light of wavelength 1.5 $\mu m$ incident on a material with a characteristic Raman frequency of $20\times 10^{12}$ Hz results in a Stokes shifted line of wavelength:
1. 1.47 $\mu m$
2. 1.57 $\mu m$
3. 1.67 $\mu m$
4. 1.77 $\mu m$

Raman shift in terms of wavelength is given by $$\frac{1}{\Delta\lambda}=\frac{1}{\lambda_0}-\frac{1}{\lambda_1}$$ $$\Delta\lambda=\frac{\lambda_0\lambda_1}{\lambda_1-\lambda_0}$$ where, $\lambda_0$ is wavelength of incident light, $\lambda_1$ is the Raman spectrum wavelength. Here, $\lambda_0=1.5\mu m$ $\lambda_1=\frac{c}{\nu}=\frac{3\times10^8}{20\times 10^{12}}=15\mu m$ $$\Delta\lambda=\frac{1.5\times15}{15-1.5}=1.67\mu m$$

Hence, answer is (C)

Problem set 65

1. What are the expected types of gamma ray transitions between the following states of odd ‘A’ nuclei : $$f_{5/2}\rightarrow P_{3/2}$$
1. E3, M4, E5, M6
2. M1, E2, M3, E4
3. M4, E5, M6, E7
4. M4, M1, E3, E4

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 $f_{5/2}\rightarrow P_{3/2}$ $$\Delta I=l=\left|\frac{5}{2}-\frac{3}{2}\right|\text{ to } \frac{5}{2}+\frac{3}{2}$$ $$\Delta I=l=1,2,3,4$$ Also, we know that the state with even $l$ value have even parity and the state with odd $l$ have odd parity. Hence, for $f_{5/2}$ parity is odd and for $P_{3/2}$ parity is odd. Hence, there is no parity change i.e. $$\Delta\pi=\text{ no}$$ For $l=1,2,3,4$ and $\Delta\pi=\text{ no}$, from above table we get that transitions as M1, E2, M3 and E4

Hence, answer is (B)

2. Which of the following molecules will not be sensitive to microwave spectroscopy?
1. $LiH$
2. $CO$
3. $CH_4$
4. $CCl_3$

Microwave spectrum can not be observed for centrosymmetric linear molecules such as $N_2$ (dinitrogen) or HCCH (ethyne), which are non-polar. Tetrahedral molecules such as $CH_4$ (methane), which have both a zero dipole moment and isotropic polarizability, would not have a pure rotation spectrum.

Hence, answer is (C)

3. Based on the additive quantum numbers such as strangeness, Baryon number, charge of the particle and Isospin, indicate the following nuclear reaction cannot be induced with the following combination: $$\pi^++n\rightarrow \pi^0+k^+$$
1. Q, B, S are conversed, but $I_3$ is not conserved
2. Q, B are conversed, but S, $I_3$ are not conserved
3. Q, $I_3$ are conversed, but B, S are not conserved
4. B, S, $I_3$ are conversed, but Q is not conserved

	$\pi^++n\rightarrow \pi^0+k^+$
Q	$1+0\rightarrow 0+1$	Conserved
B	$0+1\rightarrow 0+0 $	Not Conserved
S	$0+0\rightarrow 0+1$	Not Conserved
$I_3$	$1+\frac{1}{2}\rightarrow 1+\frac{1}{2}$	Conserved

Hence, answer is (C)

4. Ripple factor is defined as the ratio between :
1. $V_{ac}V_{dc}$
2. $V_{ac}/V_{dc}$
3. $V_{dc}/V_{ac}$
4. $V_{in}/V_{out}$

Ripple factor is defined as the ratio of the root mean square (rms) value of the ripple voltage to the absolute value of the DC component of the output voltage.

Hence, answer is (B)

5. Magnetic moment of duteron $\mu_D\ne \mu_p+\mu_n$. This is due to:
1. Spin dependence of nuclear force
2. Tensor character of nuclear force
3. Spin-orbit force part of nuclear force
4. Hard core part of the nuclear force

The deuteron consists of a neutron and a proton. The deuteron is known to have a quadrupole moment, 0.00286 barns, which tells us that the deuteron is not perfectly spherical and that the force between two nucleons is not spherically symmetric. The force between two nucleons has two components, a spherically symmetric central force and an asymmetric tensor force that depends on the angles between the spin axis of each nucleon and the line connecting.

Hence, answer is (B)