Question

A metal has work function of 2.0 eV and is illuminated by monochromatic light of wavelength 5000 \AA. Calculate (i) the threshold wavelength and (ii) the stopping potential.

Hint:

Use $E(eV) = \dfrac{1240}{\lambda (nm)}$ for quick photon energy calculation.

Answer 1

Step 1: Convert work function to joules.
\[ W = 2.0 \, eV = 2.0 \times 1.6 \times 10^{-19} J = 3.2 \times 10^{-19} J \]
Step 2: Threshold wavelength.
\[ \lambda_{th} = \frac{hc}{W} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.2 \times 10^{-19}} \] \[ \lambda_{th} \approx 6.2 \times 10^{-7} m = 6200 \, \text{\AA} \]
Step 3: Energy of incident photons.
\[ E = \frac{hc}{\lambda} = \frac{1240}{500} \, eV = 2.48 \, eV \]
Step 4: Maximum kinetic energy.
\[ K_{max} = E - W = 2.48 - 2.0 = 0.48 \, eV \]
Step 5: Stopping potential.
\[ eV_s = K_{max} \quad \Rightarrow \quad V_s = 0.48 \, V \]
Step 6: Conclusion.
(i) Threshold wavelength $\lambda_{th} = 6200 \, \text{\AA}$
(ii) Stopping potential $V_s = 0.48 \, V$