Question

The Rydberg constant for hydrogen is $10967700/m$. The shortest and longest wavelength limit in its Lyman series will be respectively

• A. $911\mathring{A}1215\mathring{A}$
• B. $1011\mathring{A}1515\mathring{A}$
• C. $711\mathring{A}575\mathring{A}$
• D. None of these

Answer 1

Answer: (A)

For Lyman series $\frac{1}{\pi }=R\left[\frac{1}{(1{)}^2}-\frac{1}{n^2}\right]$
Where, $n=2,3,4,5,?$
$\frac{1}{{\lambda }_{\text{shortest }}}=R\left[\frac{1}{(1{)}^2}-\frac{1}{\infty }\right]=10967700/s$
$\frac{1}{{\lambda }_{\text{shortest }}}=R\frac{1}{10967700}=911\mathring{A}$
Longest wavelength limit $\frac{1}{{\lambda }_{\text{shortest }}}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$
$=\frac{3}{4}R{\lambda }_{\text{longets }}=\frac{4}{3R}$
$=\frac{4}{3\times 10967700}=1215\mathring{A}$