Question

Work function of a metal surface is 3.2 eV. Find out the kinetic energy of the emitted photoelectrons in Joule, when a photon of 4.0 eV energy is incident on its surface.

Hint:

Use Einstein's photoelectric equation \(KE_{max}=E-W_0\), then convert the answer from eV to Joule using \(1\ \text{eV}=1.6\times10^{-19}\ \text{J}\).

Answer 1

Step 1: Convert every energy into Joule first.
Since \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\):
Photon energy \(E = 4.0\times1.6\times10^{-19} = 6.4\times10^{-19}\ \text{J}\).
Work function \(W_0 = 3.2\times1.6\times10^{-19} = 5.12\times10^{-19}\ \text{J}\).

Step 2: Apply conservation of energy at the surface.
The photon energy that is not used up in liberating the electron appears as its kinetic energy:
\[ KE_{max} = E - W_0 \]
Step 3: Subtract the two Joule values.
\[ KE_{max} = 6.4\times10^{-19} - 5.12\times10^{-19} \]
\[ KE_{max} = 1.28\times10^{-19}\ \text{J} \]
So the fastest photoelectrons leave the surface with a kinetic energy of about \(1.28\times10^{-19}\) J (equivalently 0.8 eV).
\[\boxed{KE_{max} = 1.28\times10^{-19}\ \text{J}}\]