If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy

Question

If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy of the emitted photoelectron is\

Professor PC Thomas Classes · Answer

According to Einstein's photoelectric equation. The kinetic energy of the emitted photoelectron is $\mathrm{K}=\mathrm{h} v-\phi_{0}$\ Where $v$ is the frequency of incident radiation and $\phi_{0}$ is a work function of the metal\ If the frequency of incident radiation is doubled, then\ $\mathrm{K}^{\prime}=2 \mathrm{~h} v-\phi_{0}-2\left(\mathrm{~h} v-\phi_{0}ight)+\phi_{0}$\ $=2 \mathrm{~K}+\phi_{0}$\ $\mathrm{K}^{\prime}>2 \mathrm{~K}$\