Question

In hydrogen spectrum the shortest wavelength in the Balmer series is ${\lambda }_0$. The largest wavelength in Lyman series for helium ion, is

• A. $\frac{5}{3}{\lambda }_0$
• B. $12{\lambda }_0$
• C. $\frac{12}{5}{\lambda }_0$
• D. $\frac{{\lambda }_0}{12}$

Answer 1

The correct option is D $\frac{{\lambda }_0}{12}$

$\frac{1}{{\lambda }_0}=R[\frac{1}{4}-\frac{1}{\infty }]$

$\frac{1}{{\lambda }_1}=R\times 4[1-\frac{1}{4}]$

$\frac{1}{{\lambda }_1}=\frac{1}{3R}$

${\lambda }_1=\frac{1}{3R}$

${\lambda }_0=\frac{4}{R}$

$3{\lambda }_1=\frac{{\lambda }_0}{4}$

${\lambda }_1=\frac{{\lambda }_0}{12}$