(a) What is the range of possible wavelengths for a neutron corresponding to a range of speeds from “thermal” at$\mathbf{\text{300K}}$(see Exercise) to$\mathbf{\text{0.01c}}$.(b) Repeat part (a), but with reference to an electron.(c) For this range of speeds, what range of dimensions D would reveal the wave nature of a neutron? Of an electron?

For the average kinetic energy of the particle$\frac{\mathbf{\text{3}}}{\mathbf{\text{2}}}{\mathbf{\text{k}}}_{\mathbf{\text{B}}}\mathbf{\text{T}}$ , the wavelength of the particle of mass m can be given as,

$\mathit{\lambda }\mathbf{\text{=}}\frac{\mathbf{\text{h}}}{\sqrt{{\mathbf{\text{3mk}}}_{\mathbf{\text{B}}}\mathbf{\text{T}}}}$…………………..(1)

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{3m}}_{\text{n}}{\text{k}}_{\text{B}}\text{T}}}\\ =\frac{{\text{6.63×10}}^{\text{-34}}\text{\hspace{0.17em}J}\cdot \text{s}}{\sqrt{\text{3}\times (\text{1.6750}\times {\text{10}}^{\text{-27}}\text{kg})\times (\text{1.38}\times {\text{10}}^{\text{-23}}J/K)\times \text{300K}}}\\ =\text{1.456}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}m}\end{array}$

Using the de-Broglie Relationship,

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\text{p}}\\ =\frac{\text{h}}{{\text{m}}_{\text{n}}\times \text{0.01c}}\\ =\frac{\text{6.63}\times {\text{10}}^{\text{-34}}\text{\hspace{0.17em}J}\cdot \text{s}}{(\text{1.67}\times {\text{10}}^{\text{-27}}\text{\hspace{0.17em}kg})\times (\text{3}\times {\text{10}}^{\text{6}}\text{m/s})}\\ =\text{1.32}\times {\text{10}}^{\text{-13}}\text{m}\end{array}$

Step 2:Analysis of mass of the electron

(b)

Do the same analysis as in part (a), but this time use the electron's mass.

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{3m}}_{\text{e}}{\text{k}}_{\text{B}}\text{T}}}\\ =\frac{\text{6.63}\times {\text{10}}^{\text{-34}}\text{\hspace{0.17em}J}\cdot \text{s}}{\sqrt{\text{3}\times (\text{9.1}\times {\text{10}}^{\text{-31}}\text{kg})\times (\text{1.38}\times {\text{10}}^{\text{-23}}J/K)\times \text{300K}}}\\ =\text{6.236}\times {\text{10}}^{\text{-9}}\text{\hspace{0.17em}m}\end{array}$

Using the de-Broglie Relationship,

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\text{p}}\\ =\frac{\text{h}}{{\text{m}}_e\times \text{0.01c}}\\ =\frac{\text{6.63}\times {\text{10}}^{\text{-34}}\text{\hspace{0.17em}J}\cdot \text{s}}{(9.1\times {\text{10}}^{\text{-31}}\text{\hspace{0.17em}kg})\times (\text{3}\times {\text{10}}^{\text{6}}\text{m/s})}\\ =\text{2.43}\times {\text{10}}^{\text{-10}}\text{m}\end{array}$

(c)

When the dimension of the slit is approximately near to their wavelength, each particle will behave as a wave, e.g. diffracted from a single slit. The neutron's wave properties will be revealed in the region of about$\text{0.1nm}$to$\text{0.0001nm}$, however, the electron will vary from$\text{0.2nm}$to $\text{6.0nm}$. You may observe that the electron range is significantly smaller than that of the neutron; this is due to the mass difference between them.

Because the electron is already traveling at high speeds at ambient temperature, the given speed isn't far from the genuine value.

Therefore,

a)$\text{1.456}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}mto\hspace{0.17em}1.32}\times {\text{10}}^{\text{-13}}\text{\hspace{0.17em}m}$

b)$\text{6.236}\times {\text{10}}^{\text{-9}}\text{\hspace{0.17em}m\hspace{0.17em}to\hspace{0.17em}2.43}\times {\text{10}}^{\text{-10}}\text{\hspace{0.17em}m}$

c) In the region of roughly, the neutron's wave characteristic will be revealed $\text{0.1nm}$to$\text{0.0001nm}$ , however, the electron will vary from$\text{0.2nm}$ to$\text{6.0nm}$