When electromagnetic radiation of wavelength 500∘A is incident on a metal sheet that has a work function of 15.6 eV, find the maximum kinetic energy of the photoelectrons emitted.
Take h=6.6×10−34J.sec
A
7.18 eV
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B
11.1 eV
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C
9.15 eV
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D
5.5 eV
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Solution
The correct option is C 9.15 eV Given λ = 500∘A=500×10−10 m
From photo-electric effect, hν=W+(KE)max ⇒hcλ=W+(K.E)max
where λ= wavelength of incident radiation W= work function of the metal sheet (KE)max= Kinetic energy of the emitted photoelectron.
So, 6.6×10−34×3×108500×10−10×1.6×10−19=15.6+(KE)max
On solving, (KE)max=9.15 eV