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

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TestHub · Accepted Answer

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

$K_{1} = h ν - Φ$

where $h$ is Planck's constant, $ν$ is the frequency of incident light, and $Φ$ is the work function of the metal.

When the frequency is doubled, the new maximum kinetic energy $K_{2}$ becomes

$K_{2} = h (2 ν) - Φ$

$K_{2} = 2 h ν - Φ$

Substituting $h ν = K_{1} + Φ$ into the equation, we get

$K_{2} = 2 (K_{1} + Φ) - Φ$

$K_{2} = 2 K_{1} + Φ$

Since the work function $Φ$ is a positive quantity, $K_{2} > 2 K_{1}$ .

Thus, the kinetic energy of the emitted photoelectrons is more than double the earlier value.

Answer: more than double

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