First slide

Photo Electric Effect

Question

The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation forwhich the stopping potential is 5 eV lies in the

Moderate

A

ultraviolet region
B

visible region
C

infrared region
D

X-ray region

Solution

According to laws of photoelectric effect
$\begin{array}{l} {KE}_{\max} = E - ϕ \\ where ϕ is work function and {KE}_{\max}, is maximum, \\ kinetic energy of photoelectron . \\ ∴ hv = {eV}_{o} + ϕ \\ or hv = 5 eV + 6 .2 eV = 11 .2 eV \\ ∴ λ = (\frac{12400}{11 .2}) \overset{o}{A} \approx 1000 \overset{o}{A} \end{array}$
Hence, the radiation lies in ultraviolet region.

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