Question

The shortest $\lambda$ for the Lyman series of hydrogen atom is:

[Given, $R_H=109678c{m}^{-1}]$

• A. $911.7A^o$
• B. $700A^o$
• C. $600A^o$
• D. $811A^o$

Answer 1

The correct option is A $911.7A^o$

For Lyman series, $n_1=1$

For shortest $\lambda$ of Lyman series; energy difference in two levels showing transition should be maximum i.e $n_2=\infty$

$\frac{1}{\lambda }$$=R_H[\frac{1}{1^2}-\frac{1}{(\infty {)}^2}]$

$\frac{1}{\lambda }$$=109678$

$\lambda$ $=911.7\times {10}^{-8}$ cm

$=911.7A^0$

Hence, option A is correct.