Question

Wavelength of first line in Lyman series is $\lambda$. The wavelength of first line in Balmer series is:

• A. $\frac{5}{27}\lambda$
• B. $\frac{32}{27}\lambda$
• C. $\frac{27}{5}\lambda$
• D. $\frac{27}{32}\lambda$

Answer 1

The correct option is C $\frac{27}{5}\lambda$

According to Bohr, the wavelength emitted when an electron jumps from $n_1^{th}$ to $n_2^{th}$ orbit is
$E=\frac{hc}{\lambda }=E_2-E_1$
$\Rightarrow \frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
For first line in Lyman series
$\frac{1}{{\lambda }_L}=R(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3R}{4}$ ......(i)
For first line in Balmer series,
$\frac{1}{{\lambda }_B}=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R}{36}$ ......(ii)
From equations (i) and (ii)
$\therefore \frac{{\lambda }_B}{{\lambda }_L}=\frac{3R}{4}\times \frac{36}{5R}=\frac{27}{5}$
$\therefore {\lambda }_B=\frac{27}{5}\lambda$ $(\because {\lambda }_L=\lambda )$