The wavelength of radiation emitted is ' λ ' when an electron jumps...

Using the Rydberg formula for the wavelength of emitted radiation:

$\dfrac{1}{\lambda} = R$\left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)

For$the transition from the second to the first orbit ( n_2 = 2$, $n_1 = 1$):

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

For$the transition from the third to the first orbit ( n_2 = 3$, $n_1 = 1$), let the wavelength be $\lambda'$:

\dfrac{1}{\lambda'} =$R \left( \dfrac{1}{1^2} - \dfrac{1}{3^2} \right) =$ R \left(1 - \dfrac{1}{9} \right) = \dfrac{8R}{9}

Dividing$the$two equations:

\dfrac{\lambda'}{\lambda} = \dfrac{\dfrac{3R}{4}}{\dfrac{8R}{9}} = \dfrac{3}{4} \times \dfrac{9}{8} = \dfrac{27}{32}

$\lambda' =$\dfrac{27}{32}\lambda

Answer: \dfrac{27}{32}\lambda