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: Write the governing law.
Einstein's photoelectric equation states that the incident photon energy is shared between the work function (energy needed to free the electron) and the maximum kinetic energy of the emitted electron:
\[ KE_{max} = E - W_0 \]where \(E\) is the energy of the incident photon and \(W_0\) is the work function of the metal.

Step 2: Substitute the given values (in eV).
\(E = 4.0\ \text{eV}\) and \(W_0 = 3.2\ \text{eV}\), so
\[ KE_{max} = 4.0\ \text{eV} - 3.2\ \text{eV} = 0.8\ \text{eV} \]
Step 3: Convert the energy from eV to Joule.
Using \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\),
\[ KE_{max} = 0.8 \times 1.6\times10^{-19}\ \text{J} \]
Step 4: Do the arithmetic.
\[ KE_{max} = 1.28\times10^{-19}\ \text{J} \]
\[\boxed{KE_{max} = 1.28\times10^{-19}\ \text{J}}\]