Question

The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the
1. X-ray region
2. Ultraviolet region
3. Visible region
4. Infrared region

Answer 1

Correct Answer - Option 2 : Ultraviolet region

CONCEPT:

• Work function (ϕ): It is the minimum amount of energy required to cause photo-emission of electrons from a metal surface when light is incident on it.
  • The work function is also known as the threshold energy.
  • The energy of the incident light is equal to the sum of the work function and the photoelectron kinetic energy.

• The kinetic energy of an emitted photoelectron is the difference in energy between incident radiation and work function and is given by 

⇒ KE = hν - ϕ₀

Where ν = frequency, ϕ₀ = Work function

EXPLANATION: 

Given - ϕ₀ = 6.2 eV and KEmax = 5 eV 

• The energy of the incident photon is given by

⇒ hν = KE + ϕ₀

Substituting the given values in the above equation

⇒ hν = 5 + 6.2 = 11.2eV   

Net incident energy/photon = 5 + 6.2 = 11.2 eV

• The wavelength of the corresponding energy can be found out as

\(⇒ E = \frac {hc}{\lambda}\)

\(⇒ \lambda = \frac{hc}{E}\)

Substituting the given values in the above equation

\(⇒ \lambda = \frac{6.626\times 10^{-34}\times 3\times 10^{8}}{11.2\times 1.6\times 10^{-19}} = 1.1\times 10^{-7} m\)

• The wavelength falls on the ultraviolet region of the electromagnetic spectrum. Hence, option 2 is the answer.