Monochromatic radiation of wavelength λ is incident on a hydrogen...

As the hydrogen atoms emit radiation of six different wavelengths, some of them must have been excited to $\text{n} = 4$, The energy in $\text{n} = 4$ state is

${E}_{4} = \dfrac{{E}_{1}}{{4}^{2}} = - \dfrac{1 3 \cdot 6 \text{e} \text{V}}{1 6} = - 0 \cdot 8 5 \text{eV.}$

The energy needed to take a hydrogen atom from its ground state to $\text{n} = 4$ is

$\text{13}\text{.6 eV - 0}\text{.85 eV = 12}\text{.75 eV}$ .

The photons of the incident radiation should have $12.75\text{eV}$ of energy. So

$\dfrac{\text{hc}}{\lambda} = 1 2 \cdot 7 5 \text{eV}$

or , $\lambda = \dfrac{\text{hc}}{1 2 \cdot 7 5 \text{eV}}$

$= \dfrac{1 2 4 2 \text{eV} \text{nm}}{1 2 \cdot 7 5 \text{eV}}$

$\text{= 97}\text{.5 nm}$.