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  4. The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is

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Q. The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is

NTA AbhyasNTA Abhyas 2022

Solution:

Let $E_{1}$ and $E_{2}$ be the KE of photoelectrons for incident light of frequency $v$ and $2v$ respectively.
Then $hv=E_{1}+\phi$ and $h2v=E_{2}+\phi_{0}$
So, $2\left(E_{1} + \left(\phi\right)_{0}\right)=E_{2}+\left(\phi\right)_{0}$ or $E_{2}=2E_{1}+\phi_{0}$
It means the KE of photoelectron becomes more than double