Also find the de-Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of $\frac{3}{2} kT$ at 300 K.

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Solution

Given,

Average Kinetic energy, E = \frac{3}{2} kT = \frac{3}{2} \times (1.38 \times 10^{- 23}) \times 300 = 6.21 \times 10^{- 21} J

(Here, k is Boltzmann constant . The value of k is 1.38 \times 10^{- 38} {JK}^{- 1})

De-broglie wavelength is given by,

λ = \frac{h}{\sqrt{2 mE}}

\Rightarrow

λ = \frac{6.63 \times 10^{- 34}}{\sqrt{2 \times 6.21 \times 10^{- 21} \times 1.675 \times 10^{- 27}}}

i.e.,

λ = 1.45 \times 10^{- 10} m .

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