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प्रश्न
A small piece of cesium metal (φ = 1.9 eV) is kept at a distance of 20 cm from a large metal plate with a charge density of 1.0 × 10−9 C m−2 on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.
(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:-
Charge density of the metal plate, `σ`= 1.0 × 10−9 Cm−2
Work function of the cesium metal, φ = 1.9 eV
Wavelength of monochromatic light, `lambda = 400 "nm" = 400 xx 10^-9 "m"`
Distance between the metal plates, d = 20 cm = 0.20 m
Electric potential due to a charged plate,
`V = E xx d`,
where E, the electric field due to the charged plate, is `σ/∈_0` and d is the separation between the plates.
`therefore V = σ/∈_0 xx d`
`= (1 xx 10^-9 xx 20)/(8.85 xx 10^-12 xx 100)` `(therefore ε_0 = 8.65 xx 10^-12 "C"^2 "N"^-1 - "m"^-2)`
= 22.598 V = 22.6 V
From Einstein's photoelectric equation,
`eV_0 = hv - W_0`
`= (hc)/lambda - W`
On substituting the respective values, we get :-
`V_0 = (4.14 xx 10^-15 xx 3 xx 10^8)/(4 xx 10^-7) - 1.9`
`= 3.105 - 1.9 = 1.205 "eV"`
= 1.205 V
As V0 is much less than 'V', the minimum energy required to reach the charged plate must be equal to 22.7eV.
For maximum KE, 'V' must have an accelerating value.
Hence maximum kinetic energy,
`K.E. = V_0 + V = 1.205 + 22.6`
= 23.8005 eV
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