Work function of a metallic surface is 5.01 eV. When a light of wavelength 2000Å falls on the metallic surface, it emits photo electrons. The potential difference required to stop the fastest photo electron is? (in Volt)
Show Hint
Use $E = \frac{12400}{\lambda\ (\text{in \AA})}$ to get photon energy instantly in eV. Subtract the work function, and the numerical value left is your stopping voltage!
Step 1: Photoelectric equation. The fastest electron's kinetic energy is $K_{max} = E - \phi$, where $E$ is the photon energy and $\phi$ is the work function.
Step 2: Stopping potential link. The stopping potential stops the fastest electron, so $K_{max} = eV_{0}$. In eV, $V_{0}$ (in volts) equals $K_{max}$ (in eV).
Step 3: Photon energy short formula. \[ E = \frac{12400}{\lambda\ (\text{in \AA})} \] Step 4: Find the photon energy. With $\lambda = 2000$ A: \[ E = \frac{12400}{2000} = 6.2 \text{ eV} \] Step 5: Subtract the work function. \[ K_{max} = 6.2 - 5.01 = 1.19 \approx 1.2 \text{ eV} \] Step 6: Read the stopping potential. So $V_{0} = 1.2$ V, which is option 1. \[ \boxed{V_{0} = 1.2 \text{ V}} \]