Problem set 86

1. A function $f(x)$ satisfies the differential equation $\frac{d^2f}{dx^2}-\omega^2f=-\delta(x-a)$, where $\omega$ is positive. The Fourier transform $\tilde{f}(k)=\int_{-\infty}^{\infty}dx\:e^{ikx}f(x)$ of $f$, and the solution of the equation are, respectively,
1. $\frac{e^{ika}}{k^2+\omega^2}$ and $\frac{1}{2\omega}\left(e^{-\omega|x-a|}+e^{\omega|x-a|}\right)$
2. $\frac{e^{ika}}{k^2+\omega^2}$ and $\frac{1}{2\omega}e^{-\omega|x-a|}$
3. $\frac{e^{ika}}{k^2-\omega^2}$ and $\frac{1}{2\omega}\left(e^{-i\omega|x-a|}+e^{i\omega|x-a|}\right)$
4. $\frac{e^{ika}}{k^2-\omega^2}$ and $\frac{1}{2i\omega}\left(e^{-i\omega|x-a|}-e^{i\omega|x-a|}\right)$
$$\frac{d^2f(x)}{dx^2}-\omega^2f(x)=-\delta(x-a)$$ Taking Fourier transform $$F\!\!\left[\frac{d^2f(x)}{dx^2}\right]\!\!-\!\!F\!\left[\omega^2f(x)\right]\!\!=\!\!-F\!\left[\!\delta(\!x\!-\!a\!)\!\right]$$ $$F\left[\frac{d^2f(x)}{dx^2}\right]=-k^2f(k)$$ $$F\left[\delta(x-a)\right]=e^{ika}$$ $$-k^2f(k)-\omega^2f(k)=-e^{ika}$$ Hence, Fourier transform of $f(x)$ is $$f(k)=\frac{e^{ika}}{k^2+\omega^2}$$ Taking Inverse Fourier transform $$f(x)=F^{-1}\left[\frac{e^{ika}}{k^2+\omega^2}\right]$$ $$f(x)=\frac{1}{2\omega}F^{-1}\left[e^{ika}\frac{2\omega}{k^2+\omega^2}\right]$$ $$f(x)=\!\!\int F^{-1}\left[\frac{e^{ika}}{2\omega}\right]F^{-1}\left[\frac{2\omega}{k^2+\omega^2}\right]$$ $$F^{-1}\left[e^{ika}\right]=\delta(x-a)$$ $$F^{-1}\left[\frac{2\omega}{k^2+\omega^2}\right]=e^{-\omega |x|}$$ Using Convolution theorem, $$f(x)=\frac{1}{2\omega}\int \delta(k-a)e^{-\omega |x-k|}dk$$ $$f(x)=\frac{1}{2\omega}e^{-\omega |x-a|}$$

Hence, answer is (B)

2. Let $E_s$ denote the contribution of the surface energy per nucleon in the liquid drop model. The ratio $E_s\left(^{27}_{13}Al\right):E_s\left(^{64}_{30}Al\right)$ is
1. 2:3
2. 4:3
3. 5:3
4. 3:2

Surface energy $$E_s\propto A^{2/3}$$ Surface energy per nucleon is $$E_s\propto \frac{A^{2/3}}{A}=\frac{1}{A^{1/3}}$$ $$\frac{E_s\left(^{27}_{13}Al\right)}{E_s\left(^{64}_{30}Al\right)}=\frac{64^{1/3}}{27^{1/3}}=\frac{4}{3}$$

Hence, answer is (B)

3. According to the shell model, the nuclear magnetic moment of the $^{27}_{13}Al$ nucleus is (Given that for a proton $g_l=1$, $g_s=5.586$, and for a neutron $g_l=0$, $g_s=-3.826$)
1. $-1.913\mu_N$
2. $14.414\mu_N$
3. $4.793\mu_N$
4. 0

In $^{27}_{13}Al$ there are odd numbers of protons (13) and even numbers of neutrons (14). The unpaired proton is responsible for magnetic moment. $^{27}_{13}Al$ has an odd proton in the $j = 5/2 1d (l=2)$ state. Magnetic moment for single nucleon is given by $$\mu_j=\left(g_ll+\frac{1}{2}g_s\right)\mu_N$$ $$\mu_j=\left(1\times2+\frac{1}{2}5.586\right)\mu_N$$ $$\mu_j=4.793\mu_N$$

Hence, answer is (C)

4. The ground state electronic configuration of $^{22}Ti$ is $[Ar]3d^54s^2$. Which state, in the standard spectroscopic notations, is not possible in this configuration?
1. $^1F_{3}$
2. $^1S_{0}$
3. $^1D_{2}$
4. $^3P_{0}$

The rules governing the term symbol for the ground state according to L-S coupling scheme are given below:

1. If all shells and subshells are full then the term symbol is $^1S_0$.
2. The spin multiplicity is maximized i.e., the electrons occupy degenerate orbitals so as to retain parallel spins as long as possible (Hund’s rule).
3. The orbital angular momentum is also maximized i.e., the orbitals are filled with highest positive $m_l$ values first.
4. The overall $S$ is calculated by multiplying $\frac{1}{2}$ times the number of unpaired electrons. The overall $L$ is calculated by adding the $m_l$ values for each electron (so if there are two electrons in the same orbital, add twice that orbital's $m_l$).
5. If the sub-shell is less than half-filled, $J = L– S$ and if the sub-shell is more than half – filled, $J = L +S$.

The term symbol is given by $^{2S+1}L_J$. The left-hand superscript of the term is the spin multiplicity, given by $2S+1$ and the right- hand subscript is given by $J$.
The electron configuration of Ti is $1s^22s^22p^63s^23p^63d^24s^2$. For $d^2$ configuration we have

	$\uparrow$	$\uparrow$
$m_l=$	+2	+1	0	-1	-2

Hence, $$\begin{align*} L&=(+2)+(+1)=3\\ \end{align*}$$ i.e. $F$ state $$\begin{align*} S&=(+1/2)+(+1/2)=1\\ \end{align*}$$ $$2S+1=3$$ $$J=L-S=3-1=2$$ Term symbol for Ti is $^3F_{2}$

Let us determine which states are possible

If we have

	$\uparrow\downarrow$
$m_l=$	+2	+1	0	-1	-2

We get $^1S_0$ state

If we have

		$\uparrow\downarrow$
$m_l=$	+2	+1	0	-1	-2

We get $^1D_2$ state

If we have

	$\uparrow$			$\uparrow$
$m_l=$	+2	+1	0	-1	-2

We get $^3P_0$ state

For $^1F_3$ state we must have $S=0$, $L=3$ and $J=3$. However, it is not possible to arrange electrons to get $S=0$, $L=3$ and $J=3$.

Hence, answer is (A)

5. The band energy of an electron in a crystal for a particular $k$-direction has the form $\epsilon(k)=A-B\cos{2ka}$, where $A$ and $B$ are positive constants and $0 < ka <\pi$. The electron has a hole-like behaviour over the following range of $k$:
1. $\frac{\pi}{4} < ka < \frac{3\pi}{4}$
2. $\frac{\pi}{2} < ka < \pi$
3. $0 < ka < \frac{\pi}{4}$
4. $\frac{\pi}{2} < ka < \frac{3\pi}{4}$

The electron has a hole-like behaviour when it has negative effective mass. $$m*=\frac{\hbar^2}{\frac{\partial^2\epsilon}{\partial k^2}}=4a^2\cos{2ka}$$ Clearly, $m*$ will be negative in the range $\frac{\pi}{4} < ka < \frac{3\pi}{4}$

Hence, answer is (A)