The threshold wavelength of a photosensitive material is equal to the frequency of $\mathrm{H}_{\mathrm{a}}$…
The threshold wavelength of a photosensitive material is equal to the frequency of $\mathrm{H}_{\mathrm{a}}$ line of hydrogen. If a photon whose frequency equal to the frequency of $\mathrm{H}_{\mathrm{b}}$ line of hydrogen is incident on this photosensitive material, the maximum kinetic energy of the emitted photoelectrons is (R-Rydberg's constant, h- Flanck's constant and c- speed of light in vacuum)
$\mathrm{Rhc}$
$\frac{5 \mathrm{Rhc}}{144}$
$\frac{7 \mathrm{Rhc}}{144}$
$\frac{\mathrm{Rhc}}{36}$
Solution
$\mathrm{H}_\alpha$ line of Balmer series
$
\begin{aligned}
& \mathrm{n}_2=3 \text { to } \mathrm{n}_1=2 \\
& \frac{1}{\lambda_0}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\mathrm{R}\left[\frac{9-4}{36}\right]
\end{aligned}
$
$\frac{1}{\lambda_0}=\frac{5 \mathrm{R}}{36}$ Here $\lambda_0$ is the threshold frequency.
$\mathrm{H}_\alpha$ line of Balmer series
$
\begin{aligned}
& \mathrm{n}_2=4 \text { to } \mathrm{n}_1=2 \\
& \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\mathrm{R}\left(\frac{1}{4}-\frac{1}{16}\right) \\
& =\mathrm{R}\left(\frac{4-1}{16}\right)=\frac{3 \mathrm{R}}{16}
\end{aligned}
$
Maximum, kinetic energy, K.E $=\frac{h c}{\lambda}-\frac{h c}{\lambda_0}$
$
=\mathrm{hc}\left[\frac{3 \mathrm{R}}{16}-\frac{5 \mathrm{R}}{36}\right]=\frac{7 \mathrm{Rhc}}{144}
$