In a hydrogen atom,if an electron transitions | vedclass

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Question

(C) The total 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

Vedclass · Accepted Answer

$5$

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(C) The total 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: $N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.Given that the electron transitions to the ground state $(n_1 = 1)$ and the total number of lines $N = 10$,we have:$10 = \frac{(n_2 - 1)(n_2 - 1 + 1)}{2}$$10 = \frac{(n_2 - 1)(n_2)}{2}$$20 = n_2^2 - n_2$$n_2^2 - n_2 - 20 = 0$Solving the quadratic equation: $(n_2 - 5)(n_2 + 4) = 0$.Since $n_2$ must be positive,$n_2 = 5$.

In a hydrogen atom,if an electron transitions from an $n$-th orbit to the ground state,and the total number of spectral lines observed is $10$,then the value of $n$ is:

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