An electron in a hydrogen atom,after absorbin | vedclass

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(C) The number of spectral lines emitted when an electron transitions from an excited state $n_2$ to a lower state $n_1$ is given by the formula $N =

Vedclass · Accepted Answer

$n_2 = 4, n_1 = 2$

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(C) The number of spectral lines emitted when an electron transitions from an excited state $n_2$ to a lower state $n_1$ is given by the formula $N = \frac{(n_2 - n_1 + 1)(n_2 - n_1)}{2}$.Given that $N = 6$,we have $\frac{(n_2 - n_1 + 1)(n_2 - n_1)}{2} = 6$,which simplifies to $(n_2 - n_1 + 1)(n_2 - n_1) = 12$.Let $x = n_2 - n_1$. Then $(x+1)x = 12$,which gives $x^2 + x - 12 = 0$. Solving this quadratic equation,we get $(x+4)(x-3) = 0$,so $x = 3$.Thus,$n_2 - n_1 = 3$. Possible pairs are $(4, 1)$ or $(5, 2)$.However,the problem states that the energy of the emitted photons can be equal to,less than,or greater than the absorbed photon energy. The absorbed photon energy corresponds to the transition from $n_1$ to $n_2$. If $n_1 = 2$ and $n_2 = 4$,the transitions include $4 \rightarrow 3, 4 \rightarrow 2, 4 \rightarrow 1, 3 \rightarrow 2, 3 \rightarrow 1, 2 \rightarrow 1$. Here,$E_{4 \rightarrow 2}$ equals the absorbed energy,$E_{4 \rightarrow 3}$ is less than the absorbed energy,and $E_{4 \rightarrow 1}, E_{3 \rightarrow 1}, E_{2 \rightarrow 1}$ are greater than the absorbed energy. This satisfies all conditions. Therefore,$n_1 = 2$ and $n_2 = 4$.

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