In a hydrogen atom,the electron is in the n^{ | vedclass

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(A) The number of spectral lines emitted when an electron transitions from an energy level $n_2$ to a lower energy level $n_1$ is given by the formula

Vedclass · Accepted Answer

$6$

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(A) The number of spectral lines emitted when an electron transitions from an energy level $n_2$ to a lower energy level $n_1$ is given by the formula $N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.Here,the electron transitions from the $n^{th}$ excited state to the second excited state.The $n^{th}$ excited state corresponds to the principal quantum number $n_2 = n + 1$.The second excited state corresponds to the principal quantum number $n_1 = 3$.Given that the number of wavelengths emitted is $10$,we have:$10 = \frac{((n+1) - 3)((n+1) - 3 + 1)}{2}$$10 = \frac{(n-2)(n-1)}{2}$$20 = n^2 - 3n + 2$$n^2 - 3n - 18 = 0$$(n-6)(n+3) = 0$Since $n$ must be positive,we get $n = 6$.

In a hydrogen atom,the electron is in the $n^{th}$ excited state. It may come down to the second excited state by emitting ten different wavelengths. What is the value of $n$?

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