Question

Every series of hydrogen spectrum has an upper end and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to $18752\dot{A}$ is

• A. Balmer series
• B. Lyman series
• C. Paschen series
• D. Pfund series

Answer 1

The correct option is C Paschen series

The formula for wavelength corresponding to transition from the state $n_2$ to state $n_1$ of a hydrogen atom is given by,

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\Rightarrow \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{R\lambda }$

$\Rightarrow \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{(1.097\times {10}^7\times 18752\times {10}^{-10})}$

$\Rightarrow \frac{1}{n_1^2}-\frac{1}{n_2^2}=0.0486=\frac{7}{144}$

The upper limit of wavelength corresponds to the transition from the successive higher energy to the final energy state of a particular spectral series.

So, If we substitute $n_1=3$ and $n_2=4$

$\Rightarrow (\frac{1}{3^2}-\frac{1}{4^2})=\frac{7}{144}$

So, final state $n_1=3$ corresponds to Paschen series.

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

Why this question? This question is based on the line spectra of hydrogen atom