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Question: (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be \[1.40 \times ...

(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40×10−10m1.40 \times {10^{ - 10}}m ?
(b) Find the de Broglie wavelength of a neutron in thermal equilibrium with matter having an average kinetic energy of (32) kT\left( {\dfrac{3}{2}} \right){\text{ }}kT at 300K300K .

Explanation

Solution

Each and every matter has dual nature, that is there is a wave-particle duality. De Broglie states that there is a wave associated with each and every matter. Microscopic particles like electrons, neutrons have long waves while macroscopic particles like car, humans have shorter waves which cannot be seen easily.

Complete answer:
De Broglie states that there is a wave associated with each and every matter. Microscopic particles like electrons, neutrons have long waves while macroscopic particles like car, humans have shorter waves which cannot be seen easily.
(a) The de Broglie wavelength and the kinetic energy can be related as follows:
Formula for kinetic energy is:
K = \dfrac{1}{2}m{v^2}\\_\\_\\_\\_(1)
Where, K is the kinetic energy, m is the mass of the object and v is its velocity.
Formula for de Broglie wavelength is:
\lambda = \dfrac{h}{{mv}}\\_\\_\\_\\_\\_(2)
On using equation second in first, we get
K=12(mh2λ2m2)K = \dfrac{1}{2}\left( {\dfrac{{m{h^2}}}{{{\lambda ^2}{m^2}}}} \right)
On simplifying the above equation, we get
K=12(h2λ2m)K = \dfrac{1}{2}\left( {\dfrac{{{h^2}}}{{{\lambda ^2}{m^{}}}}} \right)
The mass of a neutron is 1.66×10−27Kg1.66 \times {10^{ - 27}}Kg .
Planck’s constant, h is 6.6×10−34Js6.6 \times {10^{ - 34}}Js .
Put these values in above equation:
K=12((6.6×10−34Js)2(1.40×10−10m)2×1.66×10−27Kg)K = \dfrac{1}{2}\left( {\dfrac{{{{(6.6 \times {{10}^{ - 34}}Js)}^2}}}{{{{(1.40 \times {{10}^{ - 10}}m)}^2} \times 1.66 \times {{10}^{ - 27}}Kg}}} \right)
On solving, we get
K=6.75×10−21JK = 6.75 \times {10^{ - 21}}J
Therefore the kinetic energy is found to be 6.75×10−21J6.75 \times {10^{ - 21}}J .
(b) Average kinetic energy, K’ is given as:
K′=32kTK' = \dfrac{3}{2}kT
Where, k is the Boltzmann constant with value k=1.38×10−23kgm2s−2K−1  k = 1.38 \times {10^{ - 23}}kg{m^2}{s^{ - 2}}{K^{ - 1}}\;
T is temperature which is given as 300K300K .
Put these values in the above formula, we get
K′=32×1.38×10−23kgm2s−2K−1×300KK' = \dfrac{3}{2} \times 1.38 \times {10^{ - 23}}kg{m^2}{s^{ - 2}}{K^{ - 1}} \times 300K
K′=6.21×10−21JK' = 6.21 \times {10^{ - 21}}J
Now, relation between de Broglie wavelength and the average kinetic energy is:
λ′=h(2K′m)\lambda ' = \dfrac{h}{{\sqrt {(2K'm)} }}
Put the values in above equation, we get
λ′=6.6×10−34(2×6.21×10−21×1.66×10−27)\lambda ' = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{\sqrt {(2 \times 6.21 \times {{10}^{ - 21}} \times 1.66 \times {{10}^{ - 27}})} }}
λ′=1.46×10−10m\lambda ' = 1.46 \times {10^{ - 10}}m
Therefore, the de Broglie wavelength is 1.46×10−10m1.46 \times {10^{ - 10}}m .

Note:
We should note that each and every particle has a wave nature which means a wave is always associated with each particle. This is known as wave-particle duality nature. Even though we cannot see the waves associated with the macroscopic particles with our eyes, they are still there. They are not visible because the mass is large.