Answer
(a) $E = 0.060~eV$
(b) $\Delta V = 0.060~V$
(c) Photons with a wavelength of 5 nm have a high amount of energy, which is significantly higher than photons of visible light.
Work Step by Step
(a) We can find the kinetic energy of the electrons:
$\lambda = \frac{h}{p}$
$\lambda = \frac{h}{\sqrt{2~m~E}}$
$\sqrt{2~m~E} = \frac{h}{\lambda}$
$2~m~E = \frac{h^2}{\lambda^2}$
$E = \frac{h^2}{2~m~\lambda^2}$
$E = \frac{(6.626\times 10^{-34}~J~s)^2}{(2)(9.1\times 10^{-31}~kg)(5.0\times 10^{-9}~m)^2}$
$E = (9.649\times 10^{-21}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})$
$E = 0.060~eV$
(b) We can find the potential difference $\Delta V$:
$q~\Delta V = E$
$\Delta V = \frac{E}{q}$
$\Delta V = \frac{9.649\times 10^{-21}~J}{1.6\times 10^{-19}~C}$
$\Delta V = 0.060~V$
(c) Photons with a wavelength of 5 nm have a high amount of energy, which is significantly higher than photons of visible light.
