Problem set 94

1. The energy density for photons in a cavity is proportional to
1. $T^3$
2. $T$
3. $T^4$
4. $T^{4/3}$

According to Stefan Boltzmann law energy density is proportional to fourth power of temperature.

Hence, answer is (C)

2. Let $\rho$ be the density matrix for a system. Then
1. $Tr(\rho)=0$
2. $Tr(\rho) < 0$
3. $0\leq Tr(\rho) < 1$
4. $Tr(\rho)=1$

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 (D)

3. A system has only two energy levels $E_1$ and $E_2$. In equilibrium at temperature $T$, the number of particles occupying level $E_1$ is double of those occupying level $E_2$. The value of $E_2-E_1$ must be ($k$ is Boltzmann constant):
1. $kT\ln2$
2. $kT\ln3$
3. $3kT$
4. $2kT$

$$n_1\propto e^{-E_1/kT}$$ $$n_2\propto e^{-E_2/kT}$$ $$\frac{n_1}{n_2}=e{(E_2-E_1)/kT}=2$$ $$E_2-E_1=kT\ln2$$

Hence, answer is (A)

4. The quantities (i) isothermal compressibility (ii) volume coefficient of expansion are :
1. Extensive and intensive respectively
2. Intensive and extensive respectively
3. Both extensive
4. Both intensive

The properties that increase proportionally with the size of the system, such as $n$ and $V$, called extensive properties, and those defined for each small region in the system, such as $P$ and $T$, called intensive properties. Terms that are added together or are on opposite sides of an equal sign must contain the same number of extensive variables. The quotient of two extensive variables is an intensive variable.

Isothermal compressibility is given by $K=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$ and volume coefficient of expansion is given by $\alpha=-\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$.

Both of these are intensive properties.

Hence, answer is (D)

5. The chemical potential in classical limit is:
1. Zero
2. Negative
3. Positive
4. Complex quantity

Negative

Hence, answer is (B)

Problem set 93

1. Uncertainty relation holds between :
1. Time and space
2. Life time and energy
3. Position and energy
4. Momentum and energy

Life time and energy

Hence, answer is (B)

2. Addition of angular momentum $j_1=1$ and $j_2=\frac{1}{2}$ will result in 6 states, of which the number of linearly independent states with magnetic number $m=-\frac{1}{2}$ is :
1. zero
2. 6
3. 3
4. 2

For $j_1=1$ and $j_2=\frac{1}{2}$, resultant $j=j_1+j_2$ to $|j_1-j_2|$. Hence, $j=\frac{3}{2},\frac{1}{2}$ $$j=\frac{3}{2}\Rightarrow m=\frac{3}{2},\frac{1}{2},-\frac{1}{2},-\frac{3}{2}$$ $$j=\frac{1}{2}\Rightarrow m=\frac{1}{2},-\frac{1}{2}$$ Hence, there are two linearly independent states

Hence, answer is (D)

3. In a scattering event by a spherical symmetric potential, only P-wave scattering occurs. The angular distribution of differential cross-section is proportional to :
1. constant
2. $\cos\theta$
3. $\cos^2{\theta}$
4. $a+\sin\theta$

$\cos^2{\theta}$

Hence, answer is (C)

4. If energy of a two-dimensional simple harmonic oscillator $E=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac{1}{2}m\omega^2(x^2+y^2)$ is fixed to be $3\hbar\omega$, the entropy is given by ($k_B$ is Boltzmann constant):
1. $k_B\ln3$
2. $2k_B\ln3$
3. $k_B\ln2$
4. zero

The energy of a two dimensional oscillator is $$E=\left(n_x+n_y+1\right)\hbar\omega$$ The energy is fixed to be $3\hbar\omega$. Clearly, following combinations of $n_x$, $n_y$ will give energy $3\hbar\omega$
$$n_x=2, n_y=0$$ $$n_x=1, n_y=1$$ $$n_x=0, n_y=2$$ Hence, number of miscrstates accessible is $\Omega=3$ $$S=k_B\ln\Omega=k_B\ln3$$

Hence, answer is (A)

5. The equation of state for photon gas is:
1. $pV=\frac{5}{3}E$
2. $pV=\frac{2}{3}E$
3. $pV=\frac{1}{3}E$
4. $pV=\hbar\omega$ for some fixed frequency $\omega$

$pV=\frac{1}{3}E$

Hence, answer is (C)

Problem set 92

1. Given $[x_i,P_j]=i\hbar\delta_{ij}$, $i,j=1,2,3$. $[x_1,P_2^2]$ is:
1. 0
2. $i\hbar P_2$
3. $2x_1$
4. $2P_2$

\begin{align*} [x_1,P_2^2]&=P_2[x_1,P_2]+[x_1,P_2]P_2\\ &=0 \end{align*}

Hence, answer is (A)

2. Which of the following is an eigen state of square of linear momentum operator $P_x^2$?
1. $Ax^2$
2. $A\left(\sin{kx}+\cos{kx}\right)$
3. $Ae^{-\alpha x^2}$
4. $A\sin^2{kx}$

$P_x=-i\hbar\frac{\partial}{\partial x}$. Hnece, $P_x^2=-\hbar^2\frac{\partial^2}{\partial x^2}$.
Clearly, \begin{align*} {\scriptstyle P_x^2\left(A\left(\sin{kx}+\cos{kx}\right)\right)}&={\scriptstyle-\hbar^2\frac{\partial^2}{\partial x^2}\left(A\left(\sin{kx}+\cos{kx}\right)\right)}\\ &=k^2A\left(\sin{kx}+\cos{kx}\right) \end{align*}

Hence, answer is (B)

3. The electron in a hydrogen atom is in a superposition state described by the wavefunction $\psi(\vec r)=A\left[4\psi_{100}(\vec r)-2\psi_{211}(\vec r)+\sqrt{6}\psi_{210}(\vec r)-\sqrt{10}\psi_{21-1}(\vec r)\right]$, $\psi_{nlm}(\vec r)$ normalized wave function. The value of normalization constant, $A$, is:
1. $\frac{1}{3}$
2. $\frac{1}{6}$
3. $6$
4. $36$

Normalization condition is $\int\psi^*(\vec r)\psi(\vec r)dr=1$. As $\psi_{nlm}(\vec r)$ are normalized wave functions, we have, $\int\psi^*_{nlm}(\vec r)\psi_{n'l'm'}(\vec r)dr=\delta_{n'n}\delta_{l'l}\delta_{m'm}$ $$\int\psi^*(\vec r)\psi(\vec r)dr=|A|^2\left(16+4+6+10\right)=1$$ $$A=\frac{1}{6}$$

Hence, answer is (B)

4. Two coherent light sources of intensities I and 9I are used in an interference experiment. The resultant intensity at points where the waves from the two sources with phase difference $\pi$ is :
1. 16I
2. 9I
3. 4I
4. zero

Resultant intensity is given by $$I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\delta$$ $$I_R=I+9I+2\sqrt{9I}\cos\pi$$ $$I_R=10I-6I=4I$$

Hence, answer is (C)

5. Non-relativistic hydrogen atom spectrum is proportional to $-1/n^2$. The degeneracy of $n^{th}$ level is:
1. $n$
2. $2n+1$
3. $n^2$
4. $1/n^2$

The energy levels in the hydrogen atom depend only on the principal quantum number $n$. For given values of $n$ and $l$, the $( 2 l + 1 )$, states with $m_{l}=-l \rightarrow l$ are degenerate. The degree of degeneracy of the energy level $E_n$ is therefore : $\sum _{l=0}^{n-1}(2l+1)=n^{2}$, which is doubled if the spin degeneracy is included.

Hence, answer is (C)

Problem set 91

1. Which of the following equations signifies the conservative nature of the electric field $\vec E$?
1. $\vec\nabla\cdot\vec E(\vec r)=\frac{\rho(\vec r)}{\epsilon_0}$
2. $\vec\nabla\times\vec E(\vec r)=\vec 0$
3. $\vec\nabla\times\vec E(\vec r,t)=\frac{-\partial\vec B(\vec r,t)}{\partial t}$
4. $\epsilon_0\mu_0\frac{\partial\vec E(\vec r,t)}{\partial t}=\vec\nabla\times\vec B(\vec r,t)-\mu_0\vec J(\vec r,t)$

A field is said to be conservative if it can be expressed as a gradient of scalar potential $V$. The equation $\vec\nabla\times\vec E(\vec r)=\vec 0$ implies that $\vec E(\vec r)$ can be expressed as $\vec E(\vec r)=-\vec\nabla V$.

Hence, answer is (B)

2. Plane electromagnetic wave is propagating through a perfect dielectric material of refractive index $\frac{3}{2}$. The phase difference between the fields $\vec E$ and $\vec B$ associated with the wave passing through the material is
1. Zero
2. $\pi$
3. $\frac{3}{2}\pi$
4. any non-zero value between $-\pi$ and $\pi$

For a perfect dielectric the phase difference between $\vec E$ and $\vec B$ is zero.

Hence, answer is (A)

3. An electromagnetic wave is propagating in a dielectric medium of permittivity $\epsilon$ and permeability $\mu$ having an electric field vector $\vec E$ associated with the wave. The associated magnetic field $\vec H$ is
1. Parallel to $\vec E$ with magnitude $E\sqrt{\mu/\epsilon}$
2. Parallel to $\vec E$ with magnitude $E\sqrt{\epsilon/\mu}$
3. Perpendicular to $\vec E$ with magnitude $E\sqrt{\mu/\epsilon}$
4. Perpendicular to $\vec E$ with magnitude $E\sqrt{\epsilon/\mu}$

Perpendicular to $\vec E$ with magnitude $E\sqrt{\epsilon/\mu}$

Hence, answer is (D)

4. Power radiated by a point charge moving with constant acceleration of magnitude $\alpha$ is proportional to
1. $\alpha$
2. $\alpha^2$
3. $\alpha^{-1}$
4. $\alpha^{-2}$

The power radiated by a point charge is given by Larmor formula as $$P=\frac{\mu_0q^2\alpha^2}{6\pi c} $$

Hence, answer is (B)

5. The output of a laser has a bandwidth of $1.2\times10^{14}$ Hz. The coherence length $l_c$ of the output radiation is
1. 3.6 mm
2. 50 $\mu$m
3. 2.5 $\mu$m
4. 1.5 cm

Coherence length of laser is given by $$l_c=\frac{c}{\Delta\nu}$$ $$l_c=\frac{3\times10^8}{1.2\times10^{14}}=2.5 \mu m$$

Hence, answer is (C)