Question

Work function of silver is $4.7 \, \text{eV}$. When ultraviolet light of wavelength $100 \, \text{nm}$ is incident on it, the stopping potential obtained is $7.7 \, \text{V}$. Find out the value of the stopping potential for the wavelength of light of $200 \, \text{nm}$.

Hint:

Stopping potential depends only on photon energy minus work function.

Answer 1

Step 1: Photoelectric equation.
The maximum kinetic energy of emitted electron is: \[ K_{\max} = e V_s = h \nu - \phi \] where $\phi$ = work function, $h \nu = \dfrac{hc}{\lambda}$.
Step 2: For $\lambda = 100 \, \text{nm$.}
Energy of photon: \[ E_1 = \frac{1240}{100} \, \text{eV} = 12.4 \, \text{eV}. \] Kinetic energy: \[ K_{\max,1} = E_1 - \phi = 12.4 - 4.7 = 7.7 \, \text{eV}. \] This matches the given stopping potential (7.7 V).
Step 3: For $\lambda = 200 \, \text{nm$.}
Energy of photon: \[ E_2 = \frac{1240}{200} = 6.2 \, \text{eV}. \] Kinetic energy: \[ K_{\max,2} = E_2 - \phi = 6.2 - 4.7 = 1.5 \, \text{eV}. \]
Step 4: Stopping potential.
\[ V_s = \frac{K_{\max,2}}{e} = 1.5 \, \text{V}. \]
Step 5: Conclusion.
The stopping potential for $\lambda = 200 \, \text{nm}$ is $1.5 \, \text{V}$.