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प्रश्न
The threshold wavelength of silver is 3800Å. Calculate the maximum kinetic energy in eV of photoelectrons emitted, when ultraviolet light of wavelength 2600Å falls on it.
(Planck’s constant, h =6.63 x 1O-34J.s.,
Velocity of light in air, c = 3 x 108 m / s)
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उत्तर
Wavelength of silver, λ1 = 3800 Aº
Wavelength of ultraviolet light, λ2 = 2600 Aº
H = 6.63 × 10-34 Js
Velocity of light in air, c = 3 × 108 m/s
To calculate kinetic energy, K.E. = hv − hvº
`"K.E."="hc"[1/lambda-1/lambda_o]`
`"K.E."=19.89xx10^-19[1/2.6-1/3.8]`
`"K.E."=(2.416xx10^-19)/(1.6xx10^-19)`
K.E. = 1.51 eV
Hence, the maximum kinetic energy emitted by the photoelectron is 1.51 eV.
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