Question

Give Einstein's equation for photoelectric emission. The work function of a metal is 4.2 eV. Electrons are emitted due to light of wavelength 2000 Å. Find the stopping potential for the emitted electrons.

Hint:

Photon energy \( hc/\lambda = 12400/2000 = 6.2\ \text{eV} \); subtract the 4.2 eV work function, and the stopping potential in volts equals the leftover energy in eV.

Answer 1

Step 1: Set up the energy balance.
Einstein's equation says a photon's energy splits into the work function plus the electron's kinetic energy: \(h\nu = W + eV_0\), where \(eV_0\) is the energy needed to stop the fastest electron. Rearranged, \(eV_0 = h\nu - W\).

Step 2: Photon energy in joules.
\[E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3\times10^{8})}{2000\times10^{-10}}\]
\[E = \frac{1.98\times10^{-25}}{2\times10^{-7}} = 9.9\times10^{-19}\ \text{J}\]

Step 3: Work function in joules.
\[W = 4.2\ \text{eV} = 4.2\times1.6\times10^{-19} = 6.72\times10^{-19}\ \text{J}\]

Step 4: Solve for stopping potential.
\[eV_0 = E - W = 9.9\times10^{-19} - 6.72\times10^{-19} = 3.18\times10^{-19}\ \text{J}\]
\[V_0 = \frac{3.18\times10^{-19}}{1.6\times10^{-19}} = 1.99\ \text{V} \approx 2.0\ \text{V}\]
Working entirely in SI units gives the same result as the electron-volt shortcut.

\[\boxed{V_0 \approx 2.0\ \text{V}}\]