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Find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (32) kT at 300 K.

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प्रश्न

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.

संख्यात्मक
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उत्तर

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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अध्याय 11: Dual Nature of Radiation and Matter - Exercise [पृष्ठ ४०८]

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एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 11 Dual Nature of Radiation and Matter
Exercise | Q 11.17 (b) | पृष्ठ ४०८

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