Question

The wavelength of the first member of the Balmer series in hydrogen spectrum is $A^0$.Then the wave length (in $A^0$) of the first member of Lyman series in the same spectrum is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{{[\frac{1}{n_1^2}-\frac{1}{n_2^2}]}_1}{{[\frac{1}{n_1^2}-\frac{1}{n_2^2}]}_2}$

$\frac{{\lambda }_2}{x}=\frac{[\frac{1}{2^2}-\frac{1}{3^2}]}{[\frac{1}{1^2}-\frac{1}{2^2}]}$

${\lambda }_2$ = $x\frac{2}{36}\times \frac{4}{3}=\frac{5x}{27}$