Question

A spectral line of hydrogen with $\lambda =4938A^{\circ }$ belongs to the series:

• A. Lyman
• B. Balmer
• C. Paschen
• D. Brackett

Answer 1

Answer: (B)

Balmer

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

$\frac{1}{4938\times {10}^{-10}}=1.097\times {10}^7\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

$\frac{1}{4938\times {10}^{-3}}=\frac{1}{n_1^2}-\frac{1}{n_2^2}$

$\frac{1}{5.4169}=\frac{1}{n_1^2}-\frac{1}{n_2^2}$

For Lyman

$l{n}_1=1$

$\frac{1}{5.4169}=\frac{1}{1^2}-\frac{1}{n_2^2}$

not integer value

For Balmer

$l{n}_1=2$

$\frac{1}{5.4169}=\frac{1}{2^2}-\frac{1}{n_2^2}{n}_2=4$ integer value

Hence the answer is Balmer.

Hence, te correct option is $\text{B}$