The frequency of incident light falling on a photosensitive material is doubled, the K.E. of the emitted photoelectrons will be

Question

TestHub · Accepted Answer

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by

$K_{1} = h ν - W$

where $h ν$ is the energy of the incident light and $W$ is the work function of the material.

When the frequency of incident light is doubled, the new kinetic energy becomes

$K_{2} = h (2 ν) - W$

$K_{2} = 2 h ν - W$

Substituting $h ν = K_{1} + W$ into the equation for $K_{2}$ :

$K_{2} = 2 (K_{1} + W) - W$

$K_{2} = 2 K_{1} + W$

Since the work function $W$ is a positive quantity ( $W > 0$ ), we get

$K_{2} > 2 K_{1}$

Thus, the kinetic energy of the emitted photoelectrons will be more than two times its initial value.

Answer: more than two times its initial value.

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