If the energies of two light radiations E 1 and E 2 are 25 eV and 100 eV respectively, then their respective…

According to Plank's quantum theory, energy of the incident wave can be expressed as:

$\mathrm{E} = \frac{\mathrm{hc}}{\mathrm{\lambda }}$

Energies of different radiation are given i.e.

${\mathrm{E}}_1=25\mathrm{eV}, {\mathrm{E}}_2=100 \mathrm{eV}$

Now,

$$\begin{gathered} {\mathrm{E}}_1 = \frac{\mathrm{hc}}{{\mathrm{\lambda }}_1} \Rightarrow {\mathrm{\lambda }}_1=\frac{\mathrm{hc}}{{\mathrm{E}}_1} \\ {\mathrm{E}}_2 = \frac{\mathrm{hc}}{{\mathrm{\lambda }}_2} \Rightarrow {\mathrm{\lambda }}_2=\frac{\mathrm{hc}}{{\mathrm{E}}_2} \end{gathered}$$

So the ratio of wavelength at different energies

$$\begin{gathered} \frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2} =\frac{{\displaystyle \raisebox{1ex}{$\mathrm{hc}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{E}}_1$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{hc}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{E}}_2$}\right.}}=\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1} \\ \frac{{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_2}= \frac{100 \mathrm{eV}}{25 \mathrm{eV}}=\frac{4}{1} \\ \mathrm{Hence}, {\mathrm{\lambda }}_1 : {\mathrm{\lambda }}_2=4 : 1 \end{gathered}$$