Question

What is the wavelength of a neutron of kinetic energy $\mathbf{\text{1MeV}}$? Of kinetic energy$\mathbf{\text{20eV}}$? (The difference is important for the fission of uranium, as a neutron will be more easily absorbed when its wavelength is very large compared with the dimensions of the wavelength behaves more like a diffuse wave than like a localized particle)

Short answer

The wavelength of Neutron of kinetic energy

${\lambda }_{\text{1MeV}}=\text{2.87}\times {\text{10}}^{\text{-14}}m$

${\lambda }_{\text{20eV}}=\text{6.41}\times {\text{10}}^{\text{-12}}m$

Step-by-step solution

Step 1:Given data.

Given that the neutron's K.E. is $\text{1\hspace{0.17em}MeV}$and$20eV$

Wavelength of the neutron with the known kinetic energy

$\begin{array}{l}\mathbf{\text{KE}}\mathbf{=}\frac{{\mathbf{\text{p}}}^{\mathbf{\text{2}}}}{\mathbf{\text{2m}}}\\ \mathbf{\text{KE}}\mathbf{=}\frac{{\mathbf{\text{h}}}^{\mathbf{\text{2}}}\mathbf{/}{\mathbf{\text{λ}}}^{\mathbf{\text{2}}}}{\mathbf{\text{2m}}}\\ \mathbf{\text{λ}}\mathbf{=}\frac{\mathbf{\text{h}}}{\sqrt{\mathbf{\text{2m}}\mathbf{\times }\mathbf{\text{KE}}}}\end{array}$

Wavelength for the Kinetic Energy of20 eV .

$\begin{array}{c}\text{KE}=\text{20\hspace{0.17em}eV}\\ =\text{32}\times {\text{10}}^{\text{-19}}\text{\hspace{0.17em}J}\end{array}$

$\begin{array}{c}\text{λ}=\frac{\text{6.63}\times {\text{10}}^{\text{-34}}J\cdot s}{\sqrt{\text{2}\times (\text{1.67}\times {\text{10}}^{\text{-27}}kg)\times (\text{32}\times {\text{10}}^{\text{-19}}J)}}\\ =\text{6.41}\times {\text{10}}^{\text{-12}}\text{m}\end{array}$

Therefore, the wavelength of Neutron of kinetic energy

(a)${\lambda }_{\text{1MeV}}=\text{2.87}\times {\text{10}}^{\text{-14}}m$

(b)${\lambda }_{\text{20eV}}=\text{6.41}\times {\text{10}}^{\text{-12}}m$