Question

A hydrogen atom is in an excited state of principle quantum number $n$. It emits a photon of wavelength $\lambda$ when it returns to the ground state. The value of $n$ is $(\text{R is Rydberg constant)}$

• A. $\sqrt{\lambda R(\lambda R-1)}$
• B. $\sqrt{\frac{\lambda R-1}{\lambda R}}$
• C. $\sqrt{\frac{\lambda R}{\lambda R-1}}$
• D. $\sqrt{\lambda (R-1)}$

Answer 1

The correct option is C $\sqrt{\frac{\lambda R}{\lambda R-1}}$

We know that,

$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})...\left(1\right)$

Here, $n_1=1\text{(ground state)}$
$n_2=n\text{(excited state)}$

On substituting these values in equation $\left(1\right)$, we get,

$\Rightarrow \frac{1}{\lambda }=R(\frac{1}{1^2}-\frac{1}{n^2})$

$\Rightarrow \frac{1}{\lambda }=R(1-\frac{1}{n^2})=R\left(\frac{n^2-1}{n^2}\right)$

$\Rightarrow n^2=\lambda R(n^2-1)=\lambda R{n}^2-\lambda R$

$\Rightarrow (\lambda R-1)n^2=\lambda R$

$\Rightarrow n=\sqrt{\frac{\lambda R}{(\lambda R-1)}}$

Hence, option $\left(C\right)$ is correct.