Question

In a hydrogen spectrum,$\lambda$ be the wavelength of first transition line of Lyman series. The wavelength difference will be "a $\lambda$ " between the wavelength of $3^{\text {rd }}$ transition line of Paschen series and that of $2^{\text {nd }}$ transition line of Balmer Series where $a =$ ________

Answer 1

For first line of Lyman
$ \frac{1}{\lambda}= R \left(1-\frac{1}{4}\right)= R \left(\frac{3}{4}\right) $
$ \Rightarrow \lambda=\frac{4}{3 R} $
$ 3^{\text {rd }} \text { line(Paschen) } $
$ \frac{1}{\lambda_3}= R \left(\frac{1}{3^2}-\frac{1}{6^2}\right)=\frac{ R }{9} \times \frac{3}{4} $
2nd line(Balmer) $ \frac{1}{\lambda_2}= R \left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{ R }{4} \times \frac{3}{4} $
Thus $ a \lambda=\lambda_3-\lambda_2=\frac{12}{ R }-\frac{16}{3 R }=\frac{20}{3 R }$
putting (1) $ a\left(\frac{4}{3 R}\right)=\frac{20}{3 R} \Rightarrow a=5 $