Chapter 28, Problem 56GP

To determine

Separation of interference peaks.

Answer to Problem 56GP

  $\text{48}\text{.9}\text{nm}$

Explanation of Solution

Given info: Mass of proton, $\text{m = 1}{\text{.67×10}}^{\text{-27}}\text{kg}$

Protons are accelerated from rest across, $\text{550}\text{V}$

Distance, $\text{d}\text{=}0.\text{70}\text{mm}\text{=}\text{0}\text{.70}{\text{×10}}^{\text{-3}}\text{m}$

Length, $\text{L}\text{=}\text{28}\text{m}$

Formula used:

From the de Broglie equation of matter the wavelength is

  $\text{λ=}\frac{\text{h}}{\text{p}}\text{=}\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{0}}\left(\text{KE}\right)}}$

  $\text{λ}$ is wavelength

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

  $p$ is photon’s momentum

  $KE$ is Kinetic energy.

  $m_0$ is rest mass

Calculation:

First calculate the wavelength of the proton by using the formula $\text{λ=}\frac{\text{h}}{\text{p}}\text{=}\frac{\text{h}}{\sqrt{{\text{2m}}_{\text{0}}\left(\text{KE}\right)}}$

  $\text{=}\frac{\left(\text{6}{\text{.63×10}}^{\text{-34}}\text{J×s}\right)}{{\left[2\left(\text{1}\text{.67}\times {\text{10}}^{\text{-27}}\text{kg}\right)\left(550\text{eV}\right)\left(1.60\times {10}^{-19}\text{J/eV}\right)\right]}^{\frac{\text{1}}{\text{2}}}}\text{=}\text{1}\text{.22}\times {\text{10}}^{\text{-12}}\text{m}$

The peaks of the interference pattern are given as $\text{d}\text{sin}\text{θ =nλ,}\text{n =}\text{1,2,}\mathrm{...}$ and the positions on the screen are $\text{y}\text{=}\text{L}\text{tan}\text{θ}$

For small angles $\text{sin}\text{θ}=\text{tanθ}$

So, $\text{y=}\frac{\text{nLλ}}{\text{d}}$

Thus, the separation:

  $\begin{array}{l}\text{Δy =}\frac{\text{Lλ}}{\text{d}}\text{=}\frac{\left(\text{28}\text{m}\right)\left(\text{1}{\text{.22×10}}^{\text{-12}}\text{m}\right)}{\text{0}\text{.70}{\text{×10}}^{\text{-3}}\text{m}}\\ \text{=}\text{4}{\text{.89×10}}^{\text{-8}}\text{m }\\ \text{=48}\text{.9}\text{nm}\end{array}$

Conclusion:

The required value is $\text{48}\text{.9}\text{nm}$ .