Question

The stopping potential for the photoelectric current is obtained as 4 V for a metal surface by a monochromatic light of wavelength \( \lambda \), incident on it. The stopping potential becomes \(V\) for the incident light of wavelength \( 2\lambda \) on this surface. Find the formula of threshold wavelength \( \lambda_0 \) in terms of \( \lambda \).

Hint:

In the photoelectric effect, the stopping potential is related to the energy of the incident photons. The threshold wavelength is the maximum wavelength of light that can cause the emission of photoelectrons from the surface.

Answer 1

Photoelectric Effect and Stopping Potential:
The photoelectric equation, which describes the energy conservation for the photoelectric effect, is given by: \[ E_{\text{photon}} = E_{\text{kinetic}} + E_{\text{work}}, \] where: - \( E_{\text{photon}} \) is the energy of the incident photon,
- \( E_{\text{kinetic}} \) is the kinetic energy of the emitted electron, and
- \( E_{\text{work}} \) is the work function, which is the minimum energy required to release an electron from the surface. The energy of a photon is given by: \[ E_{\text{photon}} = h\nu = \frac{hc}{\lambda}, \] where: - \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light, and
- \( \lambda \) is the wavelength of the incident light. The stopping potential \(V\) is related to the kinetic energy of the emitted electron: \[ E_{\text{kinetic}} = eV, \] where \( e \) is the charge of the electron, and \( V \) is the stopping potential. The work function is \( W = h\nu_0 \), where \( \nu_0 \) is the threshold frequency for the photoelectric effect. For the incident light of wavelength \( \lambda \), the stopping potential is 4V, so we can write: \[ h\left( \frac{c}{\lambda} \right) = e . 4 + h\nu_0. \] For the light of wavelength \( 2\lambda \), the stopping potential becomes \( V \), so we have: \[ h\left( \frac{c}{2\lambda} \right) = eV + h\nu_0. \] By subtracting these two equations, we get: \[ h \left( \frac{c}{\lambda} - \frac{c}{2\lambda} \right) = e(4 - V). \] Simplifying the left-hand side: \[ h . \frac{c}{2\lambda} = e(4 - V), \] and solving for \( V \), we get: \[ V = 4 - \frac{hc}{2e\lambda}. \] Now, for the threshold wavelength \( \lambda_0 \), the stopping potential \( V = 0 \). At this point, the photon energy is equal to the work function: \[ h . \frac{c}{\lambda_0} = h . \nu_0. \] Thus, the formula for the threshold wavelength \( \lambda_0 \) in terms of \( \lambda \) is: \[ \lambda_0 = 2 \lambda. \] This shows that the threshold wavelength is twice the wavelength for which the stopping potential is 4V.