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Question
In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (φ = 4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector.
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Solution
Given:-
Work function of copper, ϕ = 4.5 eV,
Wavelength of monochromatic light, λ = 200 nm
From Einstein's photoelectric equation, kinetic energy,
`K = E - phi = (hc)/lambda - phi`,
where, h = Planck's constant
c = speed of light
`therefore K = 1242/200 - 4.5`
`= 6.21 - 4.5 = 1.71 "eV"`
Thus, at least 1.7 eV is required to stop the electron. Therefore, minimum kinetic energy will be 2 eV.
It is given that electric potential of 2 V is required to accelerate the electron. Therefore, maximum kinetic energy
`= (2 + 1.7) "eV" = 3.7 "eV"`
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