When photons of energy hv fall on a photosensitive surface of work function $E_0$, photoelectrons of maximum…
When photons of energy hv fall on a photosensitive surface of work function $E_0$, photoelectrons of maximum energy $k$ are emitted. If the frequency of radiation is doubled the maximum kinetic energy will be equal to ( $\mathrm{h}=$ Planck's constant)
k
2 k
$\mathrm{k}+\mathrm{E}_0$
$\mathrm{k}+\mathrm{h} v$
Solution
In photoelectric effect,
$\begin{aligned}
\text { Total energy } & =K \cdot E \text { of electrons }+ \text { Work function } \\
h v & =k+w_0 \\
\therefore w_0 & =h v-k
\end{aligned}$
If frequency is doubled, let max. $k \cdot E$ be $k^{\prime}$
$\begin{aligned}
\therefore \quad h \times 2 v & =k^{\prime}+w_0 \\
\therefore 2 h v & =k^{\prime}+(h v-k) \\
\therefore \quad k^{\prime} & =h v+k
\end{aligned}$