Question

When light of wavelength \( \lambda \) is incident on a photosensitive surface, the stopping potential is \( V \). When light of wavelength \( 3\lambda \) is incident on the same surface, the stopping potential is \( \frac{V}{6} \). Threshold wavelength for the surface is

• A. \( 5\lambda \)
• B. \( 9\lambda \)
• C. \( 3\lambda \)
• D. \( 6\lambda \)

Hint:

In the photoelectric effect, the stopping potential depends on the frequency (or wavelength) of the incident light and can be used to find the threshold wavelength.

Answer 1

The Correct Option is A

Step 1: Use the photoelectric equation.
The photoelectric equation is given by: \[ eV = h \nu - \phi \] where \( e \) is the charge of the electron, \( V \) is the stopping potential, \( h \) is Planck’s constant, \( \nu \) is the frequency of the light, and \( \phi \) is the work function of the material. The frequency \( \nu \) is related to the wavelength \( \lambda \) by: \[ \nu = \frac{c}{\lambda} \] where \( c \) is the speed of light. Step 2: Calculate the change in stopping potential.
For the light of wavelength \( \lambda \), the stopping potential is \( V \). For light of wavelength \( 3\lambda \), the stopping potential is \( \frac{V}{6} \). Using the photoelectric equation, we get: \[ eV = h \frac{c}{\lambda} - \phi \] \[ e \frac{V}{6} = h \frac{c}{3\lambda} - \phi \] By solving these two equations, we can find that the threshold wavelength is \( 5\lambda \). Step 3: Conclusion.
Thus, the threshold wavelength for the surface is \( 5\lambda \), which corresponds to option (A).