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Question
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
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Solution
Kinetic energy of the electron, Ek = 120 eV
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
(a) For the electron, we can write the relation for kinetic energy as:
`"E"_"k" = 1/2"mv"^2`
Where,
v = Speed of the electron
∴ `"v"^2 = sqrt((2"eE"_"k")/"m")`
= `sqrt((2 xx 1.6 xx 10^(-19) xx 120)/(9.1 xx 10^(-31)))`
= `sqrt(42.198 xx 10^12)`
= 6.496 × 106 m/s
Momentum of the electron, p = mv
= 9.1 × 10−31 × 6.496 × 106
= 5.91 × 10−24 kg m s−1
Therefore, the momentum of the electron is 5.91 × 10−24 kg m s−1.
(b) Speed of the electron, v = 6.496 × 106 m/s
(c) De Broglie wavelength of an electron having a momentum p, is given as:
`lambda = "h"/"p"`
= `(6.6 xx 10^(-34))/(5.91 xx 10^(-24))`
= 1.116 × 10−10 m
= 0.112 nm
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
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