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

Answer: 5

For first line of Lyman
$\frac{1}{\lambda }=R\left(1-\frac{1}{4}\right)=R\left(\frac{3}{4}\right)$
$\Rightarrow \lambda =\frac{4}{3R}$
$3^{\text{rd }}\text{ line(Paschen) }$
$\frac{1}{{\lambda }_3}=R\left(\frac{1}{3^2}-\frac{1}{6^2}\right)=\frac{R}{9}\times \frac{3}{4}$
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 $a\lambda ={\lambda }_3-{\lambda }_2=\frac{12}{R}-\frac{16}{3R}=\frac{20}{3R}$
putting (1) $a\left(\frac{4}{3R}\right)=\frac{20}{3R}\Rightarrow a=5$