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Question
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me = 9.11 × 10−31 kg).
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Solution
An X-ray probe has greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ = 1 Å = 10−10 m
Mass of an electron, me = 9.11 × 10−31 kg
Planck’s constant, h = 6.6 × 10−34 Js
Charge on an electron, e = 1.6 × 10−19 C
The kinetic energy of the electron is given as:
`"E" = 1/2 "m"_"e""v"^2`
`"m"_"e""v" = sqrt(2"Em"_"e")`
Where,
v = Velocity of the electron
mev = Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
`lambda = "h"/"p" = "h"/("m"_"e""v") = "h"/sqrt(2"Em"_"e")`
∴ E = `"h"^2/(2lambda^2"m"_"e")`
`= (6.6 xx 10^(-34))^2/(2xx(10^(-10))^2 xx 9.11 xx 10^(-31))`
= 2.39 × 10−17 J
= `(2.39 xx 10^(-17))/(1.6 xx 10^(-19))`
= 149.375 eV
Energy of a photon `"E'" = "hc"/(lambda"e") "eV"`
= `(6.6 xx 10^(-34) xx 3 xx 10^8) /(10^(-10) xx 1.6 xx 10^(-19))`
= 12.375 × 103 eV
= 12.375 keV
Hence, a photon has greater energy than an electron for the same wavelength.
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