The $\lambda$ of $\mathrm{H}_{\alpha}$ line of Balmer series is $6500 Å$. What is the $\lambda$ of…

The $\lambda$ of $\mathrm{H}_{\alpha}$ line of Balmer series is $6500 Å$. What is the $\lambda$ of $\mathrm{H}_{\beta}$ line of Balmer series?
  1. $4814.8 Å$
  2. $5814.8 Å$
  3. $2867.4 Å$
  4. $9395.3 Å$

Solution

For $\mathrm{H}_{\alpha}$ lines of Balmer series $\mathrm{n}_{1}=2, \mathrm{n}_{2}=3$
For $\mathrm{H}_{\beta}$ line of Balmer series $\mathrm{n}_{1}=2, \mathrm{n}_{2}=4$
$\therefore \quad \frac{1}{\lambda_{\mathrm{H}_{\alpha}}}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}ight]$ ...(1)
and $\frac{1}{\lambda_{\mathrm{H} \beta}}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}ight]$ ...(2)
By Eqs. (1) and (2)
$\frac{\lambda_{\beta}}{\lambda_{\alpha}}=\frac{\frac{1}{4}-\frac{1}{9}}{\frac{1}{4}-\frac{1}{16}} \quad \therefore \quad \lambda_{\beta}=\lambda_{\alpha} \times\left[\frac{80}{108}ight]=6500 \times \frac{80}{108}=4814.8 Å$ /

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more STRUCTURE OF ATOM questions on Aicharya