Question

The frequency and work function of an incident photon are \[\nu \] and \[{\phi _0}\] ​. If \[{\nu _0}\] ​ is the threshold frequency the necessary condition for the emission of photoelectron is
A. \[\nu < {\nu _0}\]
B. \[\nu = \dfrac{{{\nu _0}}}{2}\]
C. \[\nu \ge {\nu _0}\]
D. None of these

Answer 1

Hint: The required condition for the emission of photoelectron is that the energy of the incident photon be sufficient enough to overcome the attractive force which keeps the electron of the atom fixed in the orbit of the atom. So, the minimum energy of the photon is equal to the work function of the metal.

Formula used:
\[K = h\nu - {\phi _0}\]
where K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\nu \]is the frequency of the photon and \[\phi \]is the work function of the metal.

Complete step by step solution:
When the photon is incident on metal then it contains energy which is able to overcome the attractive force with which the valence electron is bound to the shell of the atom of the metal. For the minimum condition, the kinetic energy of the ejected electron is zero. Using the energy formula,
\[{\phi _0} = h{\nu _0} \ldots \left( i \right)\]

Putting in the formula of the kinetic energy, we get
\[K = h\nu - h{\nu _0}\]
\[\Rightarrow K = h\left( {\nu - {\nu _0}} \right)\]
For the emission of the electron, the kinetic energy of the ejected electron must be non-zero. If the energy of the photon exceeds the minimum energy needed to eject the electron then the rest of the energy is transferred as kinetic energy of the ejected electrons. So,
\[K \ge 0\]
\[\Rightarrow h\left( {\nu - {\nu _0}} \right) \ge 0\]
\[\Rightarrow \nu - {\nu _0} \ge 0\]
\[\therefore \nu \ge {\nu _0}\]
Hence, the frequency of the incident light must not be less than the threshold frequency of the metal.

Therefore, the correct option is C.

Note: As the energy of the photon is proportional to the frequency of the photon, so if the work function increases then the minimum energy of the photon for the photo emission should also increase, i.e. the frequency should increase.