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)