Question

A hydrogen atom emits a photon of wavelength $\frac{36}{35R}$ 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, ${\lambda }_{n^{\text{th }}}=\frac{36}{35R}$
As wavelength in lyman-series,
$\frac{1}{{\lambda }_{Ly}}=R\left[1-\frac{1}{n^2}\right](\because n=2,3,4)$
$\Rightarrow {\lambda }_{Ly}=\frac{n^2}{R\left(n^2-1\right)}$
Where, ${\lambda }_{Ly}$ is the wavelength of the photon, when a electron jumps from $n^{\text{th }}$ state to ground state.
Hence, ${\lambda }_{Ly}={\lambda }_{n^{\text{th }}}$
$\Rightarrow \frac{n^2}{R\left(n^2-1\right)}=\frac{36}{35R}$
$\Rightarrow n^2=36$
$\Rightarrow n=6^{\text{th }}$ state.