Question

The first line of Balmer series has wavelength $6563\dot{\text{A}}$. What will be the wavelength of the first line of Lyman series?

• A. $1215.4\dot{\text{A}}$
• B. $6563\dot{\text{A}}$
• C. $7500\dot{\text{A}}$
• D. $2500\dot{\text{A}}$

Answer 1

The correct option is A $1215.4\dot{\text{A}}$

We know that,

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

The first line of Balmer series corresponds to the transition from $n_2=3\to n_1=2$

$\Rightarrow \frac{1}{{\lambda }_B}=R(\frac{1}{2^2}-\frac{1}{3^2})$

$\Rightarrow \frac{1}{{\lambda }_B}=R(\frac{1}{4}-\frac{1}{9})$

$\Rightarrow \frac{1}{{\lambda }_B}=\frac{5R}{36}.....\left(1\right)$

Similarly,
The first line of Lyman series corresponds to the transition from $n_2=2\to n_1=1$

$\Rightarrow \frac{1}{{\lambda }_L}=R(\frac{1}{1^2}-\frac{1}{2^2})$

$\Rightarrow \frac{1}{{\lambda }_L}=\frac{3R}{4}.....\left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow \frac{1/{\lambda }_B}{1/{\lambda }_L}=\frac{\left(\frac{5R}{36}\right)}{\left(\frac{3R}{4}\right)}=\frac{5}{3}\times \frac{1}{9}$

$\Rightarrow \frac{{\lambda }_L}{{\lambda }_B}=\frac{5}{27}$

$\Rightarrow {\lambda }_L=\frac{5}{27}{\lambda }_B=\frac{5}{27}\times 6563\dot{\text{A}}$

$\Rightarrow {\lambda }_L=1215.4\dot{\text{A}}$

Hence, option $\left(\text{A}\right)$ is correct.

Why this question? This question is very useful to understand the emission spectra of hydrogen atom, which very important concept for JEE Mains.