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
-
ultraviolet region
-
visible region
-
infrared region
-
X-ray region
Solution
A.
ultraviolet region
According to laws of photoelectric effect
KEmax = E - Φ
where Φ is work function and KEmax is the maximum kinetic energy of photoelectron.
hv = eVo + Φ
or
hv = 5eV + 6.2 eV = 11.2 eV
therefore, λ = (12400/11.2) = 1000 A
Hence, the radiation lies in ultraviolet region.
