Question

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

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

Hint:

In photoelectric effect problems, the stopping potential decreases as the wavelength of the incident light increases.

Answer 1

The Correct Option is B

Step 1: Understanding the photoelectric effect.
The energy of a photon is given by: \[ E = \frac{hc}{\lambda} \] Where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. The stopping potential \( V \) is related to the energy of the photon by: \[ eV = \text{Energy of photon} - \text{Work function of metal} \] Step 2: Using the new wavelength.
For the new wavelength \( 2\lambda_0 \), the energy of the photon is halved, and the stopping potential decreases accordingly. Therefore, the new stopping potential will be: \[ V_0 - \frac{hc}{2e\lambda_0} \] Thus, the correct answer is (B) \( V_0 - \frac{hc}{2e\lambda_0} \).