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Question
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
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Solution
Wavelength of an electron (`lambda_"e"`) and a photon `(lambda_"p"), lambda_"e" = lambda_"p" = lambda` = 1 nm = 1 × 10−9 m
Planck’s constant, h = 6.63 × 10−34 Js
(a) The momentum of an elementary particle is given by de Broglie relation:
`lambda = "h"/"p"`
`"p" = "h"/lambda`
It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.
∴ p = `(6.63 xx 10^(-34))/(1 xx 10^(-9))`
= 6.63 × 10−25 kg ms−1
(b) The energy of a photon is given by the relation:
`"E" = "hc"/lambda`
Where,
Speed of light, c = 3 × 108 m/s
∴ E = `(6.63 xx 10^(-34) xx 3 xx 10^8)/(1 xx 10^(-9) xx 1.6 xx 10^(-19)`
= 1243.1 eV
= 1.243 keV
Therefore, the energy of the photon is 1.243 keV.
(c) The kinetic energy (K) of an electron having momentum p,is given by the relation:
`"K" = 1/2 "p"^2/"m"`
Where,
m = Mass of the electron = 9.1 × 10−31 kg
p = 6.63 × 10−25 kg ms−1
∴ K = `1/2 xx (6.63 xx 10^(-25))^2/(9.1 xx 10^(-31))`
= 2.415 × 10−19 J
= `(2.415 xx 10^(-19))/(1.6 xx 10^(-19))`
= 1.51 eV
Hence, the kinetic energy of the electron is 1.51 eV.
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