Question

In hydrogen spectrum, the shortest wavelength in Balmer series is $\lambda$. The shortest wavelength in Brackett series will be

• A. $2\lambda$
• B. $4\lambda$
• C. $9\lambda$
• D. $16\lambda$

Answer 1

Answer: (B)

$\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
For shortest wavelength in Balmer series,
$n_1=2;n_2=\infty$
$\frac{1}{\lambda }=R\left[\frac{1}{2^2}-\frac{1}{{\infty }^2}\right]$ or $\lambda =\frac{4}{R}$
For shortest wavelength in Brackett series,
$\frac{1}{{\lambda }^{\prime }}=R\left[\frac{1}{4^2}-\frac{1}{{\infty }^2}\right]$
$\Rightarrow {\lambda }^{\prime }=\frac{16}{R}=4\times \frac{4}{R}=4\lambda$