Question

Whenever a photom is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavwlength does this latter photon correspond to ?

Answer 1

The second wave length is form Balmer to lyman i,e. from

n = 2 to n = 1

$n_1=2,n_2=1$

$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$\frac{1}{\lambda }=1.097\times {10}^{-7}[\frac{1}{(2{)}^2}-\frac{1}{(1{)}^2}]$

$\frac{1}{\lambda }=1.097\times {10}^{-7}[\frac{1}{4}-1]$

= $-1.097\times \frac{3}{4}\times {10}^7$

$\lambda =\frac{4}{1.097\times 3}\times {10}^{-7}$

= $1.215\times {10}^{-7}$

= $121.5\times {10}^{-9}=122nm$