In a hydrogen spectrum λ be the wavelength of first transition line...

Using Rydberg formula,

$\dfrac{1}{\lambda} = R\left(1{n}_{f}^{2} - \dfrac{1}{{n}_{i}^{2}}\right)$, where, $R$ is Rydberg constant and $n$ is quantum number.

For first line of Lyman series

$\dfrac{1}{\lambda} = R\left(1 - \dfrac{1}{4}\right) = R\left(\dfrac{3}{4}\right)$

$⇒ \lambda = \dfrac{4}{3R} . . . \left(1\right)$

For ${3}^{rd}$ line of Paschen series

$\dfrac{1}{{\lambda}_{3}} = R\left(\dfrac{1}{{3}^{2}} - \dfrac{1}{{6}^{2}}\right) = \dfrac{R}{9} \times \dfrac{3}{4}$

For ${2}^{nd}$ line of Balmer series

$\dfrac{1}{{\lambda}_{2}} = R\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{4}^{2}}\right) = \dfrac{R}{4} \times \dfrac{3}{4}$

Thus, $a\lambda = {\lambda}_{3} - {\lambda}_{2} = \dfrac{12}{R} - \dfrac{16}{3R} = \dfrac{20}{3R}$

Putting equation \left(1\right)$</p><p>$a\left(\dfrac{4}{3R}\right) = \dfrac{20}{3R} ⇒ a = 5