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)