When photons of energy $h v$ fall on an aluminium plate (of work function $E_0$ ), photoelectrons of maximum…
When photons of energy $h v$ fall on an aluminium plate (of work function $E_0$ ), photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be:
$K+h v$
$K+E_0$
$2 K$
$K$
Solution
According to given situation
$\begin{aligned} & h v & =E_0+K \\ \text { and } & 2 h v & =E_0+k^{\prime} \\ \Rightarrow & k^{\prime} & =k+h v\end{aligned}$