Question

When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$, then stopping potential is $3V_s$. When the same surface is illuminated by light of wavelength $2\lambda$, then stopping potential becomes $V_s$. The value of threshold wavelength for photoelectric emission will be:

• A. $4\lambda$
• B. $8\lambda$
• C. $4\lambda/3$
• D. $6\lambda$

Hint:

In the photoelectric effect, the stopping potential is related to the wavelength of incident light, and the threshold wavelength can be determined using the relation involving stopping potentials.

Answer 1

The Correct Option is D

The stopping potential $V_s$ is related to the energy of the photons by the equation: \[ eV_s = h \left( \frac{c}{\lambda} - \frac{c}{\lambda_0} \right) \] where: $\lambda_0$ is the threshold wavelength, $h$ is Planck's constant, and $c$ is the speed of light. For $\lambda$, the stopping potential is $3V_s$, and for $2\lambda$, it is $V_s$. Solving this equation will give the threshold wavelength as $6\lambda$.