Question

The shortest and the longest wavelength in Lyman series of hydrogen spectrum are:
$\text{Rydberg constant},R_H=109678c{m}^{-1}$

• A. 911.7 $\stackrel{o}{A}$ and 1215.7 $\stackrel{o}{A}$
• B. 566.4 $\stackrel{o}{A}$ and 788.6 $\stackrel{o}{A}$
• C. 1015.6 $\stackrel{o}{A}$ and 1438.8 $\stackrel{o}{A}$
• D. 863.1 $\stackrel{o}{A}$ and 1215.7 $\stackrel{o}{A}$

Answer 1

The correct option is A 911.7 $\stackrel{o}{A}$ and 1215.7 $\stackrel{o}{A}$

For Lyman series, $n_1=1.$
For the shortest wavelength in Lyman series (i.e., series limit), the energy difference in the two states showing the transition should be maximum, i.e., $n_2=\infty$
So, $\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{{\infty }^2}]=R_H\Rightarrow \lambda =\frac{1}{109678}=9.117\times {10}^{-6}cm=911.7\stackrel{o}{A}$

For the longest wavelength in Lyman series (i.e., first line), the energy difference between the states showing the transition should be minimum, i.e. $n_2=2$
$\frac{1}{\lambda }=R_H[\frac{1}{1^2}-\frac{1}{2^2}]=\frac{3}{4}{R}_H\Rightarrow \lambda =\frac{4}{3}\times \frac{1}{R_H}=\frac{4}{3\times 109678}=1215.7\times {10}^{-8}cm=1215.7\stackrel{o}{A}$