Question

The wavelength of the first line of Balmer series is $6563\mathring{\text{A}}$. The wavelength of the first line of Lyman series will be

• A. $1215\mathring{\text{A}}$
• B. $2500\mathring{\text{A}}$
• C. $7500\mathring{\text{A}}$
• D. $600\mathring{\text{A}}$

Answer 1

The correct option is A $1215\mathring{\text{A}}$

Given,

Wavelength of first line of Balmer series $\left({\lambda }_1\right)=6563\mathring{\text{A}}$

As we know that the wavelength of emitted photon from hydrogen atom is given by,

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

The first line of Balmer series is given by,

$\frac{1}{{\lambda }_1}=R[\frac{1}{2^2}-\frac{1}{3^2}]=R(\frac{1}{4}-\frac{1}{9})=\frac{5R}{36}.....\left(1\right)$

The first line of Lyman series is given by,

$\frac{1}{{\lambda }_2}=R[\frac{1}{1}-\frac{1}{4}]=\frac{3R}{4}.....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right),$

$\therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{\frac{5}{36}}{\frac{3}{4}}=\frac{5}{27}$

$\Rightarrow {\lambda }_2=\frac{5}{27}{\lambda }_1=\frac{5}{27}\times 6563\approx 1215\mathring{\text{A}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.