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NCERT Class XII Chapter
Dual Nature of Radiation and Matter
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Question : 17 of 37
Marks: +1, -0
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−1010^{-10} m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter having an average kinetic energy of 32\frac{3}{2} kT at 300 K.
Solution:  
(a) de Broglie wavelength, λ = h2mK\frac{h}{\sqrt{2mK}}
∴ Kinetic energy (K) of neutron
K = h22mλ2\frac{h^2}{2m\lambda^2} =
(6.63×10−34)22×1.677×10−27×(1.40×10−10)2\frac{(6.63 \times 10^{-34})^2}{2 \times 1.677 \times 10^{-27} \times (1.40 \times 10^{-10})^2}
= 6.686 × 10−2110^{-21} J
= 6.686×10−211.6×10−19\frac{6.686 \times 10^{-21}}{1.6 \times 10^{-19}} eV = 4.178 × 10−210^{-2} eV
(b) K = 32\frac{3}{2} kT, where k = Boltzmann constant
∴ λ = h2mK\frac{h}{\sqrt{2mK}} = h3mkT\frac{h}{\sqrt{3mkT}}
=
6.63×10−343×1.677×10−27×1.38×10−23×300\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.677 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}
m = 6.63×10−1020.8\frac{6.63 \times 10^{-10}}{\sqrt{20.8}} = 6.63×10−104.56\frac{6.63 \times 10^{-10}}{4.56} m = 1.45 × 10−1010^{-10} = 0.145 nm
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