Answer:
[tex]1.51\cdot 10^{-4}V[/tex]
Explanation:
The de Broglie wavelength of the electron is given by
[tex]\lambda=\frac{h}{mv}[/tex]
where
[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electron
v is its speed
We want the electron to have a wavelength of
[tex]\lambda = 0.1 \mu m = 0.1\cdot 10^{-6}m[/tex] (the diameter of the pinhole)
Substituting in the equation above, we find what speed the electron should have:
[tex]v=\frac{h}{m\lambda}=\frac{6.63\cdot 10^{-34}}{(9.11\cdot 10^{-31})(0.1\cdot 10^{-6})}=7278 m/s[/tex]
Now, when a charged particle is accelerated through a potential difference, the kinetic energy it gains is equal to the change in electric potential energy, therefore:
[tex]e\Delta V = \frac{1}{2}mv^2[/tex]
where
[tex]e=1.6\cdot 10^{-19} C[/tex] is the electron charge
[tex]\Delta V[/tex] is the potential difference
And solving for [tex]\Delta V[/tex],
[tex]\Delta V=\frac{mv^2}{2e}=\frac{(9.11\cdot 10^{-31})(7278)^2}{2(1.6\cdot 10^{-19})}=1.51\cdot 10^{-4}V[/tex]
