Question

If light photons are incident on a photosensitive metal of work function 2.1 ev and the
maximum kinetic energy of the photoelectrons emitted is 1.9ev, the energy of
incident photons is?​

Answer 1

The energy of incident photons with maximum K.E. of the photoelectrons as 1.9 eV is 4 eV.

Explanation:

The work function of the photosensitive metal, ɸ = 2.1 eV

The maximum kinetic energy of the photoelectrons emitted, K.E.max = 1.9 eV

We know that,

• The minimum energy needed for photo emission of electrons is known as Threshold Energy or Work Function. It is usually given in the form of electron-volt (eV).

• The excess energy present in the photon is gained by the electron and it becomes its Kinetic Energy.

So, the formula of the Photoelectric Effect is:

E = ɸ + K.E.max

where

E = Energy of a photon

ɸ = Work Function

K.E. max = Maximum Kinetic Energy of Electron

By substituting the given values in the above formula, we get

E = 2.1 + 1.9 = 4 eV

Thus, the energy of the incident photons is 4 eV.

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