The first member of the Paschen series in hydrogen spectrum is of wavelength 18,800 Å. The short wavelengths limit of Paschen series is
Answer & explanation
Correct answer: option 3
For Paschen series $\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{3^2}-\frac{1}{n^2}\right]$ ; n = 4, 5, 6....
For first member of Paschen series n = 4
$\frac{1}{\lambda_1}=R\left[\frac{1}{3^2}-\frac{1}{4^2}\right] \Rightarrow \frac{1}{\lambda_1}=\frac{7 R}{144}$
$\Rightarrow R=\frac{144}{7 \lambda_1}=\frac{144}{7 \times 18800 \times 10^{-10}}=1.1 \times 10^{-7}$
For shortest wave length $n=\infty$
So $\frac{1}{\lambda}=R\left[\frac{1}{3^2}-\frac{1}{\infty^2}\right]=\frac{R}{9}$
$\Rightarrow \lambda=\frac{9}{R}=\frac{9}{1.1 \times 10^{-7}}=8.225 \times 10^{-7} m=8225 ~Å$