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प्रश्न
A horizontal cesium plate (φ = 1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically-upward component of velocity is non-positive for each photoelectron?
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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उत्तर
Given :-
Work function of the cesium plate, φ = 1.9 eV
Wavelength of radiation, λ = 250 nm
Energy of a photon,
`E = (hc)/lambda`,
where h = Planck's constant
c = speed of light
`therefore E = 1240/250 = 4.96 "eV"`
From Einstein's photoelectric equation, kinetic energy of an electron,
`K = E - phi`
`⇒ K = (hc)/lambda - phi`
[Here , h is Planck's constant and c is the speed of light]
`⇒ K = 4.96 "eV" - 1.9 "eV"`
= 3.06 eV
For non-positive velocity of each photo electron, the velocity of a photoelectron should be equal to minimum velocity of the plate.
∴ Velocity of the photoelectron,
`v = sqrt((2K)/m)` (m = mass of electron)
`therefore v = sqrt((2 xx 3.06 xx 1.6 xx 10^-19)/(9.1 xx 10^-31))`
`= 1.04 xx 10^6 "ms"^-1`
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