Question

If the wavelength of the first line of the Balmer series in the hydrogen spectrum is $\lambda$, then the wavelength of the first line of the Lyman series is

• A. $\left(\frac{27}{5}\right)\lambda$
• B. $\left(\frac{5}{27}\right)\lambda$
• C. $\left(\frac{32}{27}\right)\lambda$
• D. $\left(\frac{27}{32}\right)\lambda$

Answer 1

The correct option is B

$\left(\frac{5}{27}\right)\lambda$

For first line of Balmer series

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

$\Rightarrow R=\frac{36}{5\lambda }$

$\therefore$ Wave length of the first line, ${\lambda }_L$ of the Lyman series is given by

$\frac{1}{{\lambda }_L}=R(1-\frac{1}{4})=\frac{36}{5\lambda }\times \frac{3}{4}=\frac{27}{5\lambda }$

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