Problem set 6

1. If the electrostatic potential were given by $\phi=\phi_0\left(x^2+y^2+z^2\right)$, where $\phi_0$ is constant, then the charge density giving rise to the above potential would be
1. 0
2. $-6\phi_0\epsilon_0$
3. $-2\phi_0\epsilon_0$
4. $-\frac{6\phi_0}{\epsilon_0}$
$$\begin{align*} \bar E&=-\bar \nabla\phi\\ &=-\bar \nabla\left\{\phi_0\left(x^2+y^2+z^2\right)\right\}\\ &=-2\phi_0(x\hat i+y\hat j+z\hat k) \end{align*}$$ Charge density $\sigma=\epsilon_0\bar\nabla.\bar E=-6\phi_0\epsilon_0$
2. A set of 15 distinguishable particles are placed in 3 energy states such that 2 particles in the first state, 12 in the second state and 1 in the third state. The number of distinct arrangements are:
1. 1365
2. 15
3. 455
4. $3^{15}$
Number of ways of arranging $N$ distinguishable particles into groups of $N_j$ each, such that $N_1$ in state 1, $N_2$ in state 2, etc. and $N_j$ in the $j^{th}$ state are given by $$W(N)=\frac{N!}{N_1!N_2!\dots N_j!}$$ $$W(15)=\frac{15!}{2!\times12!\times1!}=1365$$
3. The critical temperature for the Bose-Einstein condensation depends on the density of particles as :
   1. $n^{1/3}$
   2. $n^{2/3}$
   3. $n$
   4. $n^{1/2}$
   $N=\frac{V(2\pi mkT)^{3/2}}{h^3}2.162$. Hence (B)$n^{2/3}$
4. In $^3S$ state of the Helium atom, the possible values of the total electronic angular momentum quantum numbers are:
   1. 0 (zero) only
   2. 1 only
   3. 0,1 and 2
   4. 0 and 1 only
   The term symbol is given by $^{(2s+1)}L_J$. For $^3S$ state $(2s+1)=3$, hence $s=1$. For $S$ state $L=0$. Hence $J=L+S=1$. Hence answer is (B)1 only
5. The probability that two friends have the same birth month is:
1. $\frac{1}{6}$
2. $\frac{1}{12}$
3. $\frac{1}{36}$
4. $\frac{1}{144}$
A person can have birthday in any of the 12 months of the year. Another person can also have his/her birthday in any of the 12 months. Now,the total number of possible combinations is 12*12 , which equals 144. This set of all possible combinations contains 12 pairs of identical months {for example(Jan,Jan)}, which are our favourable outcomes.
Thus the probability =
(Number of favourable outcomes)/(Number of events in sample space)=12/(12*12)=1/12