Light of frequency $4 v_{0}$ is incident on the metal of the threshold frequency $v_{0}$ . The maximum kinetic energy of the emitted photoelectrons is

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Light of frequency ν falls on material of threshold frequency v0. Maximum kinetic energy of emitted electron is proportional to:

The correct option is A v−v0By photoelectric equation;⇒hν=hν0+KE⇒KE=h(ν−ν0)Clearly kinetic energy is directly proportional to h(ν−ν0)Hence, option A is correct.

Light of frequency $ν$ falls on material of threshold frequency $v_{0}$ . Maximum kinetic energy of emitted electron is proportional to:

The correct option is A $v - v_{0}$
By photoelectric equation;
$\Rightarrow h ν = h ν_{0} + K E \Rightarrow K E = h (ν - ν_{0})$

Clearly kinetic energy is directly proportional to $h (ν - ν_{0})$
Hence, option A is correct.