Question

In a photoelectric effect measurement, the stopping potential for a given metal is found to be \( V_0 \) volt when radiation of wavelength \( \lambda_0 \) is used. If radiation of wavelength \( 2\lambda_0 \) is used with the same metal then the stopping potential (in volt) will be:

• A. \( \dfrac{V_0}{2} \)
• B. \( 2V_0 \)
• C. \( V_0 + \dfrac{hc}{2e\lambda_0} \)
• D. \( V_0 - \dfrac{hc}{2e\lambda_0} \)

Hint:

The stopping potential in the photoelectric effect depends on the energy of the incident photons, which is inversely proportional to the wavelength.

Answer 1

The Correct Option is D

Step 1: The stopping potential is related to the energy of the incoming photons by the equation: \[ E = h\nu - eV_0, \] where \( \nu = \dfrac{c}{\lambda_0} \) is the frequency of the radiation.
Step 2: For wavelength \( 2\lambda_0 \), the energy of the photons is half of the energy for \( \lambda_0 \). Therefore, the stopping potential decreases, and the new stopping potential is: \[ V_0' = V_0 - \dfrac{hc}{2e\lambda_0}. \]
Final Answer: \[ \boxed{V_0 - \dfrac{hc}{2e\lambda_0}} \]