A hydrogen atom falls from $\mathrm{n}^{ ext {th }}$ higher energy orbit to first energy orbit ( $\mathrm{n}=1$ ). The energy released is equal to 12.75 ev . The $n^{ ext {th }}$ orbit is

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A hydrogen atom falls from $\mathrm{n}^{	ext {th }}$ higher energy orbit to first energy orbit ( $\mathrm{n}=1$ ). The energy released is equal to 12.75 ev . The $n^{	ext {th }}$ orbit is

Aicharya · Accepted Answer

In hydrogen atom, $\begin{aligned} & \Delta \mathrm{E}=13.6\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}ight) \ & \Rightarrow 12.75=13.6\left(\frac{1}{1^2}-\frac{1}{\mathrm{n}^2}ight) \ & 	herefore \mathrm{n}=4 .\end{aligned}$

A hydrogen atom falls from $\mathrm{n}^{\text {th }}$ higher energy orbit to first energy orbit (…

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