Problem set 7

1. For a face centred cubic crystalline structure the following diffraction peaaks may be observed in the X-ray diffraction experiment:
1. (110), (330)
2. (111), (321)
3. (100), (321)
4. (111), (331)
For FCC structure- The peaks for which all h,k, l are even or all h,k, l are odd are present.
For BCC structure- The peaks for which h+k+l=even, are present.
For SCstructure- All peaks are present
Hence answer is (D)(111), (331)
2. In the X-ray diffraction of a set of crystal planes having $'d'$ equal to 0.18 nm, a first order reflection is found to be at an angle of $12^o$. The wavelength of X-ray used is [Given: $\sin{12^o}=0.2079$]
1. 0.1543 nm
2. 0.0749 nm
3. 0.0374 nm
4. 0.749 nm
Bragg's law is $$2d\sin\theta=n\lambda$$ For first order pattern $n=1$ $$\begin{align*} \lambda&=2\times0.18\times10^{-9}\sin12^o\\ &=2\times0.18\times10^{-9}\times0.2079\\ &=0.0749 nm \end{align*}$$
3. The outer electron configuration of divalent Manganese ion is $3d^54s^0$. The ground state of this ion is characterized by the spectroscopic term:
1. $^6S_{5/2}$
2. $^2D_{5/2}$
3. $^2F_{5/2}$
4. $^6H_{5/2}$
The rules governing the term symbol for the ground state according to L-S coupling scheme are given below:
1. 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).
2. The orbital angular momentum is also maximized i.e., the orbitals are filled with highest positive m values first.
3. 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 $J$.
For $d^5$ configuration we have
	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$
$m_l=$	+2	+1	0	-1	-2
Hence, $$\begin{align*} L&={\scriptstyle(+2)+(+1)+(0)+(-1)+(-2)}\\ &=0 \end{align*}$$ i.e. $S$ state $$\begin{align*} S&={\scriptstyle(+1/2)+(+1/2)+(+1/2)+(+1/2)+(+1/2)}\\ &=5/2 \end{align*}$$ $$2S+1=6$$ $$J=L+S=0+5/2=5/2$$ Term symbol is $^6S_{5/2}$
4. Consider a nuclear $F^{19}$. When it is placed in a magnetic field of 1.0 tesla, the resonance frequency (in units of MHz) of the signal observed for this nucleus in the NMR spectrometer is : (Given: $g_N=5.256$, $\mu_N=5.0504\times10^{-27} J/T$; the subscript $N$ refers to the nuclear factors)
1. 30 MHz
2. 90 MHz
3. 40 MHz
4. 5.0 MHz
The resonance condition for NMR is $$h\nu=g_N\mu_NB_0$$ $$\begin{align*} \nu&=\frac{g_N\mu_NB_0}{h}\\ &=\frac{5.256\times5.0504\times10^{-27}\times1}{6.626\times10^{-34}}\\ &=40~MHz \end{align*}$$
5. The work done in bringing a charge $+q$ from infinity in free space, to a position at a distance $d$ in front of a semi-infinite grounded metal surface is
1. $-\frac{q^2}{4\pi\epsilon_0(d)}$
2. $-\frac{q^2}{4\pi\epsilon_0(2d)}$
3. $-\frac{q^2}{4\pi\epsilon_0(4d)}$
4. $-\frac{q^2}{4\pi\epsilon_0(6d)}$
In front of a semi-infinite grounded metal surface when a charge $q$ is held at a distance $x$, then there is an image charge $-q$ at a distance $x$ behind the metal surface. Hence force on $+q$ charge is given by $$F=-\frac{q^2}{4\pi\epsilon_0(2x)^2}$$ Hence force required to move the charge is $$F=\frac{q^2}{4\pi\epsilon_0(2x)^2}$$ $$\begin{align*} W&=\int_{\infty}^{d}Fdx\\ &=\frac{q^2}{4\pi\epsilon_0(4)}\int_{\infty}^d\frac{1}{x^2}dx\\ &=-\frac{q^2}{4\pi\epsilon_0(4d)} \end{align*}$$