$(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;4$

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Answer : (b) 2

Explanation :

Level C must correspond to n=4 since total number of possible transitions from level 4 to lower levels is

1+2+3=6

The level B corresponds either to n=2 or n=3

The energies of all photons involved in transitions is (in eV)

$\varepsilon _{4.2}=13.6Z^2\;(\large\frac{1}{2^2}-\large\frac{1}{4^2})=2.55 Z^2$

$\varepsilon_{4,3}=0.66 Z^2$

$\varepsilon_{4,1}=12.75 Z^2$

$\varepsilon_{3,2}=1.89 Z^2$

$\varepsilon_{3,1}=12.1 Z^2$

$\varepsilon_{2,1}=10.2 Z^2$

The only possible choices are

a) Z=1 and $\;\varepsilon_{4,2}=2.55 eV \approx 2.7eV$

b) Z=1 and $\;\varepsilon_{4,3}=2.64 eV \approx 2.7eV$

The choice Z=2 however is inconsistent with the fact that some photons have energy less than 2.7eV

Thus , Z=1 and quantum number of level B is n2 .

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