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(B) Let the electron be initially in state $n_1$ and excited to state $n_2$ by the incident photon.The total number of photons emitted when returning

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$2^{nd}$ excited state

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(B) Let the electron be initially in state $n_1$ and excited to state $n_2$ by the incident photon.The total number of photons emitted when returning from $n_2$ to $n=1$ is given by $\frac{n_2(n_2-1)}{2} = 10$.Solving for $n_2$,we get $n_2^2 - n_2 - 20 = 0$,which gives $n_2 = 5$.The energy of the incident photon is $E_{photon} = E_{n_2} - E_{n_1} = E_5 - E_{n_1}$.Transitions from $n_i$ to $n_f$ result in photons with energy $E_{n_i} - E_{n_f}$.We are given that $7$ transitions have energy greater than $E_5 - E_{n_1}$.The total number of possible transitions from $n=5$ to $n=1$ is $10$. These are: $(5 	o 4, 5 	o 3, 5 	o 2, 5 	o 1, 4 	o 3, 4 	o 2, 4 	o 1, 3 	o 2, 3 	o 1, 2 	o 1)$.Transitions with energy greater than $E_5 - E_{n_1}$ are those where the final state $n_f If $n_1 = 3$,the incident photon energy is $E_5 - E_3$. Transitions with energy $> E_5 - E_3$ are $(5 	o 2, 5 	o 1, 4 	o 2, 4 	o 1, 3 	o 2, 3 	o 1, 2 	o 1)$,which are exactly $7$ transitions.Thus,$n_1 = 3$,which corresponds to the $2^{nd}$ excited state.

An electron initially present in an excited state of $H$ atom is further excited to another energy level by an incident photon. It releases $10$ photons while coming back to the ground state,out of which $7$ have higher energy than the incident photon. The electron was initially present in

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