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
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a
ultraviolet region
b
visible region
c
infrared region
d
X-ray region
answer is A.
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Detailed Solution
According to laws of photoelectric effectKEmax=E−ϕwhere ϕ is work function and KEmax, is maximum,kinetic energy of photoelectron.∴hv=eVo+ϕor hv=5eV+6.2eV=11.2eV∴λ=1240011.2Ao≈1000AoHence, the radiation lies in ultraviolet region.
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