Chapter 28, Problem 52GP

To determine

The spacing between the bright fringes of the interference pattern produced on the screen.

Answer to Problem 52GP

  $\text{1}\text{.0}\text{μm}$

Explanation of Solution

Given info:

Mass of electron, $\text{m = 9}{\text{.11×10}}^{\text{-31}}\text{kg}$

Kinetic energy, $\text{KE}\text{=}\text{45}\text{keV}\text{=}\text{45}\text{×}{\text{10}}^{\text{3}}\text{eV}$

Distance, $\text{d}\text{=}\text{2}\text{.0}{\text{×10}}^{\text{-6}}\text{m}$

Length $\text{L}\text{=}\text{35}\text{cm}\text{=}\text{0}\text{.350}\text{m}$

Formula used:

From the de Broglie equation of matter the wavelength is

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

Where,

  $\text{λ}$ 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 electron by using the formula

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

  $\begin{array}{l}\text{=}\frac{\left(\text{6}{\text{.63×10}}^{\text{-34}}\text{J×s}\right)}{{\left[2\left(\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}\right)\left(\text{45}\times {\text{10}}^{\text{3}}\text{eV}\right)\left(1.60\times {10}^{-19}\text{J/eV}\right)\right]}^{\frac{\text{1}}{\text{2}}}}\\ \text{=5}\text{.79}\times {\text{10}}^{\text{-12}}\text{m}\end{array}$

The peaks of the interference pattern are at:

  $\text{d}\text{sin}\text{θ =nλ,}\text{n =}\text{1,2,}\mathrm{...}$ and the positions on the screen are $\text{y}\text{=}\text{L}\text{tan}\text{θ}$

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

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

Thus, the separation:

  $\begin{array}{l}\text{Δy =}\frac{\text{Lλ}}{\text{d}}\text{=}\frac{\left(\text{0}\text{.350}\text{m}\right)\left(\text{5}{\text{.79×10}}^{\text{-12}}\text{m}\right)}{\text{2}\text{.0}{\text{×10}}^{\text{-6}}\text{m}}\\ \text{=}\text{1}\text{.0}{\text{×10}}^{\text{-6}}\text{m =1}\text{.0}\text{μm}\end{array}$

Conclusion:

The required value is $\text{1}\text{.0}\text{μm}$ .