Problem set 64

1. Consider the differential equation $\frac{d^2x}{dt^2}-3\frac{dx}{dt}+2x=0$. If $x=0$ at $t=0$ and $x=1$ at $t=1$, the value of $x$ at $t=2$ is
1. $e^2+1$
2. $e^2+e$
3. $e+2$
4. $2e$

Characteristic equation is $$m^2-3m+2=0$$ $$\Rightarrow (m-1)(m-2)=0$$ $$\Rightarrow m=1,2$$ Hence solution is $$x=c_1e^{t}+c_2e^{2t}$$ $x=0$ at $t=0$ $\Rightarrow c_1=-c_2$

and $x=1$ at $t=1$ $\Rightarrow c_1e+c_2e^{2}=1$ $$\Rightarrow c_1=\frac{1}{e-e^2}$$ $$c_2=-\frac{1}{e-e^2}$$ At $t=2$ $$\begin{align*} x&=c_1e^{2}+c_2e^{4}\\ &=\frac{e^{2}}{e-e^2}-\frac{e^{4}}{e-e^2}\\ &=e+e^2 \end{align*}$$

Hence, answer is (B)

2. An electric dipole of moment $\vec P$ and length $l$ aligned along the $z$-axis is used to generate electromagnetic waves. Initially it was operated at frequency 10 MHz and its power along the equatorial plane at certain distance $d (d >> l)$ was measured as $P_1$. Later, the same dipole was operated at frequency 40 MHz and power $P_2$ was measured at the same point. How do you compare the two powers ?
1. $P_2= 256 P_1$
2. $P_2= 64 P_1$
3. $P_2= 128 P_1$
4. $P_2= 16 P_1$

The average total power radiated by oscillating electric dipole is given by $$P=\frac{\mu_0\omega^4p_0^2}{12\pi c}$$ Thus, $$P\propto \omega^4$$ $$P_1\propto (10)^4$$ $$P_2\propto (40)^4$$ $$\frac{P_2}{P_1}=\left(\frac{40}{10}\right)^4$$ $$P_2= 256 P_1$$

Hence, answer is (A)

3. The number of fundamental vibrational modes of $CO_2$ molecule is :
1. 4 : 2 Raman active and 2 IR active
2. 4 : 1 Raman active and 3 IR active
3. 3 : 1 Raman active and 2 IR active
4. 3 : 2 Raman active and 1 IR active

$CO_2$ molecule is a linear molecule. Hence, number of vibrational modes are given by $3N-5$. $CO_2$ is triatomic molecule, hence $N=3$. Hence, number of modes $=3\times 3-5=4$. Out of which 1 is Raman active and 3 are IR active.

Hence, answer is (B)

4. The $L$, $S$ and $J$ quantum numbers corresponding to the ground state electronic configuration of Boron $(z = 5)$ are :
1. $L = 1$, $S = 1/2$, $J = 3/2$
2. $L = 1$, $S = 1/2$, $J = 1/2$
3. $L = 1$, $S = 3/2$, $J = 1/2$
4. $L = 0$, $S = 3/2$, $J = 3/2$

To determine ground state, we will use following Hund's rules

1. Given more than one allowed value for $S$, choose largest possible value.
2. Given more than one allowed value for $L$, choose largest possible value.
3. Given more than one allowed value for $J$, choose largest possible value of $J$ if subshell is more than half-filled and choose smallest possible value of $J$ if subshell is less than half-filled.

The ground state of Boron has a $1s^22s^22p^1$ configuration. Hence, there is single electron in $2p$ state, for which, $S=1/2$, $L=1$. Hence, $J=L-S=1/2$.

Hence, answer is (B)

5. What are the expected types of gamma ray transitions between the following states of odd ‘A’ nuclei : $$g_{9/2}\rightarrow P_{1/2}$$
1. M4 and E5
2. M1 and E2
3. M3 and E4
4. M6 and E7

Gamma rays emitted in electric $E$-$l$ transitions carry off angular momentum $l$ and parity $(-1)^l$. Therefore, angular momenta and parity of final and initial states are related by $$\Delta I=|I_i-I_f| \text{ to } I_i-I_f=l$$ $$\text{and }\Delta\pi=(-1)^l$$ The selection rules for magnetic $M$-$l$ transitions are $$\Delta I=|I_i-I_f|\text{ to } I_i-I_f=l$$ $$\text{and }\Delta\pi=(-1)^{l-1}$$ Following table shows selections rules for $\gamma$-ray transitions

Type	$l=\Delta I$	$\Delta\pi$
E1	1	yes
M1	1	no
E2	2	no
M2	2	yes
E3	3	yes
M3	3	no
E4	4	no
M4	4	yes
E5	5	yes
M5	5	no

For $g_{9/2}\rightarrow P_{1/2}$ $$\Delta I=l=\left|\frac{9}{2}-\frac{1}{2}\right|\text{ to } \frac{9}{2}+\frac{1}{2}=4, 5$$ Also, we know that the state with even $l$ value have even parity and the state with odd $l$ have odd parity. Hence, for $g_{9/2}$ parity is even and for $P_{1/2}$ parity is odd. Hence, there is parity change i.e. $$\Delta\pi=\text{ yes}$$ For $l=4,5$ and $\Delta\pi=\text{ yes}$, from above table we get that transitions as M4 and E5

Hence, answer is (A)