Answer
22 volts.
Work Step by Step
First assume that the electron is nonrelativistic. Find the wavelength using equation 27–8.
$$\lambda=\frac{h}{p}=\frac{h}{mv}$$
$$v=\frac{h}{m\lambda}=\frac{6.626\times10^{-34}J \cdot s }{(9.11\times10^{-31}kg)(0.26\times10^{-9}m)}$$
$$=2.797\times10^{6}m/s=0.0093c$$
Our assumption is valid because v is a tiny fraction of c.
Find the kinetic energy of the electron.
$$KE=\frac{1}{2}mv^2=\frac{1}{2}(9.11\times10^{-31}kg)( 2.797\times10^{6}m/s)^2$$
$$ =3.56\times10^{-18}J\approx 22 eV$$
To produce an electron with 22 eV of kinetic energy, it must be accelerated through a potential difference of 22 volts.
