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 $\frac{1}{\lambda_{\text {shortest }}}=R\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$
$=\frac{3}{4} R \lambda_{\text {longets }}=\frac{4}{3 R}$
$=\frac{4}{3 \times 10967700}=1215 \mathring{A}$