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)