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: Einstein's photoelectric equation.
When light of frequency \( \nu \) (energy \( h\nu \) per photon) falls on a metal of work function \( W \), the maximum kinetic energy of an emitted electron is
\[ K_{max} = h\nu - W = \frac{hc}{\lambda} - W \]
and the stopping potential \( V_0 \) is defined by \( eV_0 = K_{max} \).

Step 2: Energy of the incident photon.
Using \( E = \dfrac{hc}{\lambda} \) with the shortcut \( hc = 12400\ \text{eV·Å} \) and \( \lambda = 2000\ \text{Å} \):
\[ E = \frac{12400}{2000} = 6.2\ \text{eV} \]
(Check with SI: \( E = \dfrac{(6.6\times10^{-34})(3\times10^{8})}{2000\times10^{-10}} = 9.9\times10^{-19}\ \text{J} = 6.19\ \text{eV} \approx 6.2\ \text{eV} \).)

Step 3: Maximum kinetic energy.
\[ K_{max} = E - W = 6.2 - 4.2 = 2.0\ \text{eV} \]

Step 4: Stopping potential.
Since \( eV_0 = K_{max} \), when energy is expressed in eV the stopping potential in volts is numerically equal to \( K_{max} \) in eV:
\[ V_0 = \frac{K_{max}}{e} = \frac{2.0\ \text{eV}}{e} = 2.0\ \text{V} \]

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