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Question

(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h
u - \phi$

Vedclass · Accepted Answer

increases to slightly more than double.

Vedclass · Answer

(A) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h
u - \phi$, where $h$ is Planck's constant, $
u$ is the frequency of incident light, and $\phi$ is the work function of the metal.Initially, $K_{max1} = h
u - \phi$.When the frequency is doubled, the new frequency is $
u' = 2
u$.The new maximum kinetic energy is $K_{max2} = h(2
u) - \phi = 2h
u - \phi$.Since $K_{max2} = 2(h
u - \phi/2)$, and $\phi > 0$, it follows that $K_{max2} > 2(h
u - \phi) = 2K_{max1}$.Therefore, the maximum kinetic energy increases to slightly more than double the initial value.

If the frequency of incident light falling on a metallic surface is doubled, what happens to the maximum kinetic energy of the emitted photoelectrons?

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