Question

A hydrogen atom emits a photon of wavelength $\frac{36}{35 R}$ when it is jumped from its $n\text{ th}$ excited state to the ground state. Then the quantum number $n$ is ( $R$ is Rydberg constant.)

• A. 8
• B. 7
• C. 5
• D. 6

Answer 1

Answer: (D)

When a electron jump from $n^{\text {th }}$ state to ground state, then the lyman-series of hydrogen spectrum will appear.
Given wavelength, $\lambda_{n^{\text {th }}}=\frac{36}{35 R}$
As wavelength in lyman-series,
$\frac{1}{\lambda_{L y}}=R\left[1-\frac{1}{n^{2}}\right] \,\,\,(\because n=2,3,4) $
$\Rightarrow \lambda_{L y}=\frac{n^{2}}{R\left(n^{2}-1\right)}$
Where, $\lambda_{L y}$ is the wavelength of the photon, when a electron jumps from $n^{\text {th }}$ state to ground state.
Hence, $\lambda_{L y}=\lambda_{n^{\text {th }}}$
$\Rightarrow \frac{n^{2}}{R\left(n^{2}-1\right)}=\frac{36}{35 R} $
$\Rightarrow n^{2}=36 $
$ \Rightarrow n=6^{\text {th }}$ state.