Question

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $\lambda$. If $R$ is the Rydberg constant, the principal quantum number $'n'$ of the excited state is

• A. $\sqrt{\frac {\lambda }{\lambda R -1}}$
• B. $\sqrt{\frac {\lambda R^{2}}{\lambda R -1}}$
• C. $\sqrt{\frac {\lambda R}{\lambda -1}}$
• D. $\sqrt{\frac {\lambda R}{\lambda R -1}}$

Answer 1

Answer: (D)

$\sqrt{\frac {\lambda R}{\lambda R -1}}$

Answer 2

Here, $n_{f}=1, n_{i}=n$
$\therefore\,\,\, \frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right) \Rightarrow \frac{1}{\lambda}=R\left(1-\frac{1}{n^{2}}\right)\,\,\,\,\,\,\,\,\,...(i)$
or $\,\,\,\,\frac{1}{\lambda R}=1-\frac{1}{n^{2}}$ or $\frac{1}{n^{2}}=1-\frac{1}{\lambda R}$
or $\,\,\,\,\ n=\sqrt{\frac{\lambda R}{\lambda R-1}}$