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Question
For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?
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Solution
De Broglie wavelength of the neutron, λ = 1.40 × 10−10 m
Mass of a neutron, mn = 1.66 × 10−27 kg
Planck’s constant, h = 6.6 × 10−34 Js
Kinetic energy (K) and velocity (v) are related as:
`"K" = 1/2 "m"_"n""v"^2` …......(1)
De Broglie wavelength (λ) and velocity (v) are related as:
`lambda = "h"/("m"_"n""v")` ........(2)
Using equation (2) in equation (1), we get:
`"K" = 1/2 ("m"_"n""h"^2)/(lambda^2 "m"_"n"^2) = "h"^2/(2lambda^2"m"_"n")`
= `(6.63 xx 10^(-34))^2/(2xx(1.40 xx 10^(-10))^2 xx 1.66 xx 10^(-27))`
= 6.75 × 10−21 J
= `(6.75 xx 10^-21)/(1.6 xx 10^-19)`
= 4.219 × 10−2 eV
Hence, the kinetic energy of the neutron is 6.75 × 10−21 J or 4.219 × 10−2 eV.
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