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Monday, 6 February 2017
Problem set 67
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:
20 cm^{-1}
40 cm^{-1}
60 cm^{-1}
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)
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 :
4\log4
4\log2
2\log2
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)
The trivalent gadolinimum ion has seven electrons in its outer orbital. The Lande g factor for this ion is:
1
\frac{3}{2}
2
\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=8J=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)
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 :
24
36
48
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
If the temperature of a black body enclosure is tripled, the number of photons will increase by a factor of :
2
9
8
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
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