Question

A beam of electrons is accelerated across a potential of 14.60 kV before passing through two slits. The electrons form an interference pattern on a screen 4.40 m in front of the slits. The first-order maximum is 8.80 mm from the central maximum. What is the distance between the slits

Answer 1

Answer:

$5.08\times 10^{-9}\ \text{m}$

Explanation:

h = Planck constant = $6.626\times 10^{-34}\ \text{Js}$

$\lambda$ = Wavelength

m = Mass of electron = $9.11\times 10^{-31}\ \text{kg}$

$\Delta V$ = Potential difference = 14.6 kV

e = Charge of electron = $1.6\times 10^{-19}\ \text{C}$

d = Distance between slits

$\sin\theta=\dfrac{8.8\times 10^{-3}}{4.4}$

We have the relation

$\dfrac{h}{\lambda}=\sqrt{2me\Delta V}\\\Rightarrow \lambda=\dfrac{h}{\sqrt{2me\Delta V}}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 14.6\times 10^3}}\\\Rightarrow \lambda=1.016\times 10^{-11}\ \text{m}$

Wavelength is given by

$d\sin\theta=\lambda\\\Rightarrow d=\dfrac{\lambda}{\sin\theta}\\\Rightarrow d=\dfrac{1.016\times 10^{-11}}{\dfrac{8.8\times 10^{-3}}{4.4}}\\\Rightarrow d=5.08\times 10^{-9}\ \text{m}$

The distance between the slits is $5.08\times 10^{-9}\ \text{m}$.