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)