Question

Light of frequency v is incident on a certain photoelectric substance with threshold frequency \[{v_0}\]. The work function for the substance is
A. \[h\upsilon \]
B. \[h{\upsilon _0}\]
C. \[h(\upsilon - {\upsilon _0})\]
D. \[h(\upsilon + {\upsilon _0})\]

Answer 1

Hint: A light with energy \[h\upsilon \]is incident on a photosensitive surface and causes photoelectric emission. The total energy of a photon is defined as the sum of the energy utilized to eject an electron and the maximum kinetic energy of electrons. The work function of a metal is the minimum amount of energy required by an electron to eject from the metal surface.

Formula used:
Kinetic energy of photoelectrons is given as:
\[K{E_{\max }} = E - \phi \]
Where E is the energy and \[\phi \] is the work function.
Also the energy of the photon is given by the equation is given as:
\[E = h\upsilon = \dfrac{{hc}}{\lambda }\].
Where \[h\] is the Plank constant, \[\upsilon \] is the frequency of incident light, c is the speed of light and \[\lambda \] is the wavelength.
Work function is given as:
\[\phi = h{\upsilon _0}\]
Where h is planck constant and \[{\upsilon _0}\] is threshold frequency.

Complete step by step solution:
Einstein gives the phenomenon of the photoelectric effect on the basis of Plank’s theory. In the photoelectric effect, the kinetic energy of photo-electrons emitted from the metal surface E and \[\phi \] is the work function of the metal. Thus the total energy of a photon is equal to the sum of the energy utilized to eject an electron and the maximum kinetic energy of electrons. As we know the maximum kinetic energy of photoelectrons is,
\[K{E_{\max }} = E - \phi {\rm{ }}\]
\[\Rightarrow K{E_{\max }} = h\upsilon - \phi {\rm{ }}\]

For a given metal, if the threshold frequency of light is \[{\upsilon _0}\] then an amount of energy \[\phi \] of the photon of light will be spent in ejecting the electron out of the metal. The work function \[\phi \] is defined as the minimum amount of energy required to induce photoemission of electrons from a metal surface and is given as
\[\phi = h{\upsilon _0}\]
Therefore, if a metal has threshold frequency \[{v_0}\] then the work function for the substance is \[h{\upsilon _0}\].

Hence option B is the correct answer.

Note: Most of the times students make mistakes while writing the Einstein’s equation of photoelectric effect. Always remember that the energy of the incident radiation is given as the sum of kinetic energy of the photoelectron and work function of a metal.