A hydrogen atom falls from n^{text{th}} highe | vedclass

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Question

(A) The energy released during a transition in a hydrogen atom is given by the formula: $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \rig

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$n=4$

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(A) The energy released during a transition in a hydrogen atom is given by the formula: $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.Here,the final orbit $n_1 = 1$ and the initial orbit $n_2 = n$.The energy released is $\Delta E = 12.75 	ext{ eV}$.Substituting the values: $12.75 = 13.6 \left( \frac{1}{1^2} - \frac{1}{n^2} ight)$.Dividing both sides by $13.6$: $\frac{12.75}{13.6} = 1 - \frac{1}{n^2}$.$0.9375 = 1 - \frac{1}{n^2}$.$\frac{1}{n^2} = 1 - 0.9375 = 0.0625$.$n^2 = \frac{1}{0.0625} = 16$.Therefore,$n = 4$.

$A$ hydrogen atom falls from $n^{\text{th}}$ higher energy orbit to the first energy orbit $(n=1)$. The energy released is equal to $12.75 \text{ eV}$. The $n^{\text{th}}$ orbit is:

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