Question

Gold has a work function, Φ, of 5.47 ev. If a sample of gold is bombarded with 191 nm photons, what will be the velocity (in m/s) of the photo-ejected electrons?

Answer 1

According to the photo effect law the photon energy would be consumed for the ejecting of the electron and giving it some kinetic energy:

$hv=\phi +\frac{mv^2}{2};$

Reorganizing the formula gives us:

$v=\sqrt{\frac{2}{m\lambda}(hc-\phi\lambda)};$

Finally,

$v=\sqrt{\frac{2}{9.1*10^{-31} kg*1.91*10^{-7}m}(6.62*10^{-34} Js*3*10^{8} m/s-8.75*10^{-19}J*1.91*10^{-7}m)}=6.02*10^{5}m/s.$