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(B) Let the electron be excited from state $n_1$ to $n_2$ by absorbing a photon of energy $E = E_{n_2} - E_{n_1}$.When returning to the ground state $

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$2^{nd}$ excited state

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(B) Let the electron be excited from state $n_1$ to $n_2$ by absorbing a photon of energy $E = E_{n_2} - E_{n_1}$.When returning to the ground state $(n=1)$ from $n_2$,the total number of photons emitted is given by $\frac{n_2(n_2-1)}{2} = 10$.Solving for $n_2$: $n_2^2 - n_2 - 20 = 0 \implies (n_2-5)(n_2+4) = 0$. Since $n_2 > 0$,we have $n_2 = 5$.The energy of the incident photon is $E_{incident} = E_5 - E_{n_1}$.There are $7$ transitions that have higher energy than the incident photon. These transitions must be from levels $n > n_1$ to levels $n \le n_1$.The number of photons emitted with energy less than or equal to the incident photon is $10 - 7 = 3$. These correspond to transitions from levels $n \le n_1$ to $n=1$.The number of such transitions is $\frac{n_1(n_1-1)}{2} = 3$.Solving for $n_1$: $n_1^2 - n_1 - 6 = 0 \implies (n_1-3)(n_1+2) = 0$. Thus,$n_1 = 3$.Since $n_1 = 3$,the electron was initially in the $2^{nd}$ excited state ($n=3$ is 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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