Question

The work function of a metal is \( 2.5 \times 10^{-9} \) joule. If the metal is exposed to a light beam of frequency \( 6.0 \times 10^{14} \) Hz, what will be the stopping potential?

Hint:

The stopping potential is the potential required to stop the fastest emitted electron in the photoelectric effect.

Answer 1

Step 1: Einstein's Photoelectric Equation.
The photoelectric equation is given by: \[ E_k = hf - \phi \] where: - \( E_k \) is the kinetic energy of the emitted electrons, - \( h \) is Planck's constant (\( h = 6.626 \times 10^{-34} \, \text{J s} \)), - \( f \) is the frequency of the incident light, - \( \phi \) is the work function of the metal. The stopping potential is related to the kinetic energy of the emitted electrons by the equation: \[ E_k = eV_s \] where \( e \) is the charge of the electron, and \( V_s \) is the stopping potential.
Step 2: Calculating the stopping potential.
The energy of the incident photons is given by: \[ E = hf = (6.626 \times 10^{-34}) \times (6.0 \times 10^{14}) = 3.976 \times 10^{-19} \, \text{J} \] Now, using the photoelectric equation: \[ E_k = 3.976 \times 10^{-19} - 2.5 \times 10^{-19} = 1.476 \times 10^{-19} \, \text{J} \] Now, using the relation \( E_k = eV_s \): \[ V_s = \frac{E_k}{e} = \frac{1.476 \times 10^{-19}}{1.6 \times 10^{-19}} = 0.92 \, \text{V} \]
Step 3: Conclusion.
The stopping potential is \( 0.92 \, \text{V} \).