The work function of a silver surface is 4.73 eV. What would happen if the silver is hit with a photon with 2.5 eV of energy? A) One electron is released from the metal.
B) A pit forms in the metal due to the impact
C) Nothing, the photon just bounces off the surface.
D) Two electrons are released from the metal.

Nothing, the photon just bounces off the surface.

Explanation:

According to Albert Einstein, a photoelectron can only be emitted from a metal surface when the energy of the incident photon is greater than the work function of the metal.

In the scenario described in the question, the work function of the metal is greater than the energy of the photon. Hence, the photon just bounces off the metal surface without emitting any electron.

The correct option is;

C) Nothing, the photon just bounces off the surface

Explanation:

The given parameters are;

The work function of the silver surface = 4.73 eV

The amount of energy of the photon = 2.5 eV

The kinetic energy, KE of an electron when ejected from a metal surface is given as follows;

KE = h·f - B·E

Where;

hf = The energy of the photon

B·E = The work function of the metal

Whereby the energy of the photon impacting on the surface of the silver is equal to or larger than the work function of the silver surface, then the energy of the photon can be reach the electrons of the silver so as to be enable the electron to be ejected through the metal surface

However, given that the energy of the photon, 2.5 eV, is less than the work function of the silver surface, 4.73 eV, the kinetic energy of an ejected electron, KE, is less 0, therefore, no electron will be ejected form the surface of the silver and the photon just bounces off the surface.

hope you got the answer

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