If $\lambda_1$ and $\lambda_2$ are the wavelength of the photons emitted, when electrons in the $n^{ ext {th }}$ orbit of hydrogen atom fall to first excited state and ground state respectively, then the value of n is

Question

If $\lambda_1$ and $\lambda_2$ are the wavelength of the photons emitted, when electrons in the $n^{	ext {th }}$ orbit of hydrogen atom fall to first excited state and ground state respectively, then the value of n is

Aicharya · Accepted Answer

We have,
$\lambda_1=\frac{h c}{\Delta E_{n \rightarrow 2}} \text { and } \lambda_2=\frac{h c}{\Delta E_{n \rightarrow 1}}$
$\begin{aligned} \Rightarrow & \quad \lambda_1\left(\frac{13.6}{4}-\frac{13.6}{n^2}\right) =\lambda_2\left(\frac{13.6}{1}-\frac{13.6}{n^2}\right) \\ \Rightarrow & n =\sqrt{\frac{4\left(\lambda_2-\lambda_1\right)}{4 \lambda_2-\lambda_1}}\end{aligned}$

If $\lambda_1$ and $\lambda_2$ are the wavelength of the photons emitted, when electrons in the $n^{\text…

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