Question

Taking the wavelength of the first Balmer line in hydrogen spectrum $(n=3$ to $n=2)$ as $660\text{nm}$, the wavelength of the second Balmer line $(n=4$ to $n=2)$ will be -

• A. $889.2\text{nm}$
• B. $388.9\text{nm}$
• C. $642.7\text{nm}$
• D. $488.9\text{nm}$

Answer 1

The correct option is D $488.9\text{nm}$

We know that,

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For $n=3$ to $n=2$ :

$\frac{1}{660}=R{Z}^2(\frac{1}{2^2}-\frac{1}{3^2})$

$\Rightarrow R{Z}^2=\frac{3}{275}$

For $n=4$ to $n=2$ :

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{2^2}-\frac{1}{4^2})$

$\Rightarrow \frac{1}{\lambda }=\frac{3R{Z}^2}{16}=\frac{3}{16}\times \frac{3}{275}$

$\Rightarrow \lambda =488.9\text{nm}$

Hence, option $\left(D\right)$ is the correct answer.