91Ó°ÊÓ

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,

$\mathit{λ}\mathbf{\text{=}}\frac{\mathbf{\text{h}}}{\sqrt{{\mathbf{\text{3mk}}}_{\mathbf{\text{B}}}\mathbf{\text{T}}}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(1)

(a)

Use equation (1) to calculate the wavelength of the particle, such that,

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{3m}}_{\text{e}}{\text{k}}_{\text{B}}\text{T}}}\\ =\frac{\text{6.63}×{\text{10}}^{\text{-34}}\text{ }Jâ‹\dots s}{\sqrt{\text{3}×(\text{9.1}×{\text{10}}^{\text{-31}}\text{kg})×(\text{1.38}×{\text{10}}^{\text{-23}}\text{ }J/K)×\text{(295‿é)}}}\\ =\text{ 6.289 n³¾}\end{array}$

(b)

We'll do the same analysis as part (a), but this time we'll use the proton's mass.

$\begin{array}{c}\text{λ}=\frac{\text{h}}{\sqrt{{\text{3m}}_{\text{p}}{\text{k}}_{\text{B}}\text{T}}}\\ =\frac{\text{6.63}×{\text{10}}^{\text{-34}}\text{ }Jâ‹\dots s}{\sqrt{\text{3}×(\text{1.6726}×{\text{10}}^{\text{-27}}\text{kg})×(\text{1.38}×{\text{10}}^{\text{-23}}\text{ }J/K)×\text{(295‿é)}}}\\ =\text{0.146 n³¾}\end{array}$

(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. As a result, while a few nanometers will suffice for the electron to express its wave nature, the proton will require these dimensions to be decreased forty times before it can manifest its wave nature at the same temperature.

So,

Therefore

a)$\text{6.289nm}$

b)$\text{0.146nm}$

c) For the electron to express its wave nature, only a few nanometers are required; nevertheless, for the proton to manifest its wave nature at the same temperature, these dimensions must be lowered by at least forty times.