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For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m? - Physics

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प्रश्न

For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?

संख्यात्मक
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उत्तर

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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