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'}=R\left[\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right] $
$\Rightarrow \lambda'=\frac{16}{R}=4 \times \frac{4}{R}=4 \lambda$