Question

In photo electric experiment, if the wavelength of incident light on the metal changes from 200 nm to 300 nm. The decrease in stopping potential is about ($\frac{hc}{e} = 1240$ eV - nm)

• A. 2.1 V
• B. 4.2 V
• C. 3.1 V
• D. 6.2 V

Hint:

Using the shortcut $\Delta V_s \approx 1240 \Delta \left(\frac{1}{\lambda}\right)$ is highly effective for photoelectric numericals.
Always keep units in eV and nm to avoid complex conversions with Joules.

Answer 1

The Correct Option is A

Step 1: Treat stopping potential as a straight line in $1/\lambda$.
Einstein's equation $eV_s = \dfrac{hc}{\lambda} - \phi$ says $V_s$ varies linearly with $1/\lambda$, with slope $\dfrac{hc}{e} = 1240\text{ eV-nm}$. The work function $\phi$ cancels out whenever we look at a difference between two wavelengths.
Step 2: Write the drop in stopping potential as slope times the change in $1/\lambda$. \[ \Delta V_s = \frac{hc}{e}\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right) = 1240\left(\frac{1}{200}-\frac{1}{300}\right) \]
Step 3: Simplify the bracket. \[ \frac{1}{200}-\frac{1}{300} = \frac{3-2}{600} = \frac{1}{600} \]
Step 4: Multiply out. \[ \Delta V_s = \frac{1240}{600} \] \[ \boxed{\Delta V_s \approx 2.1\text{ V}} \]