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Question
Find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of `(3/2)` kT at 300 K.
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Solution
Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 × 10−23 kg m2 s−2 K−1
Average kinetic energy of the neutron:
`"K'" = 3/2 "kT"`
= `3/2 xx 1.38 xx 10^(-23) xx 300`
= 6.21 × 10−21 J
The relation for the de Broglie wavelength is given as:
`lambda"'" = "h"/(sqrt(2"K'""m"_"n"))`
Where
mn = 1.66 × 10−27 Kg
h = 6.6 × 10−34 Js
K' = 6.75 × 10−21 J
∴ `lambda"'" = (6.63 xx 10^(-34))/sqrt(2 xx 6.21 xx 10^(-21) xx 1.66 xx 10^(-27))`
= 1.46 × 10−10 m
= 0.146 nm
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
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