Question

Gold has a work function of 4.9 eV. (a) Calculate the maximum kinetic energy, in joules, of the electrons emitted when gold is illuminated with ultraviolet radiation of frequency 1.7 x 10^15 Hz. (b) What is the energy expressed in eV?
(c) What is the stopping potential for these electrons? (e = - 1.6 x 10^19 - 6.6 x 10^34Js)

Answer 1

$h\nu=\phi+E_{max}$

(a)

$E_{max}=h\nu-\phi=6.6\cdot10^{-34}\cdot1.7\cdot10^{15}-4.9\cdot1.6\cdot10^{-19}=3.38\cdot10^{-19}$ J

(b)

$E_{max}=\frac{3.38\cdot10^{-19}}{1.6\cdot10^{-19}}=2.1$ eV

(c)

$\varphi=\frac{E_{max}}{e}=\frac{3.38\cdot10^{-19}}{1.6\cdot10^{-19}}=2.1$ V