Answer:
0.15 mV
Explanation:
In order to exhibit wave nature, the de Broglie wavelength of the electron must be of the same size of the diameter of the pinhole, therefore:
[tex]\lambda=0.10 \mu m = 1.0\cdot 10^{-7} m[/tex]
The de Broglie wavelength of an electron is
[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 the electron's speed
Therefore, the electron's speed must be
[tex]v=\frac{h}{m\lambda}=\frac{6.63\cdot 10^{-34}}{(9.11\cdot 10^{-31})(1.0\cdot 10^{-7})}=7278 m/s[/tex]
When accelerated through a potential difference [tex]\Delta V[/tex], the kinetic energy gained by the electron 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}[/tex] is the magnitude of the charge of the electron
So, we can find the potential difference needed:
[tex]\Delta V=\frac{mv^2}{2e}=\frac{(9.11\cdot 10^{-31})(7278)^2}{2(1.6\cdot 10^{-19})}=1.5\cdot 10^{-4}V = 0.15 mV[/tex]
