Question

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is \(3.2\ \text{V}\). If a second light having wavelength twice of the first light is used, the stopping potential drops to \(0.7\ \text{V}\). The wavelength of the first light is ________ m.

• A. \(2.2 \times 10^{-8}\)
• B. \(3.1 \times 10^{-7}\)
• C. \(2.5 \times 10^{-7}\)
• D. \(2.9 \times 10^{-8}\)

Hint:

Stopping potential depends on frequency, not on intensity of incident light.

Answer 1

The Correct Option is C

To solve this problem, we need to use the photoelectric equation, which relates the energy of the incident photon to the kinetic energy of emitted photoelectrons.

The equation for the photoelectric effect is:

\(E_{\text{photon}} = \text{KE}_{\text{max}} + W\)

where:

• \(E_{\text{photon}} = \frac{hc}{\lambda}\) is the energy of the photon. Here, \(h\) is Planck's constant \((6.626 \times 10^{-34}\ \text{Js})\), \(c\) is the speed of light \((3 \times 10^8\ \text{m/s})\), and \(\lambda\) is the wavelength of the light.
• \(\text{KE}_{\text{max}} = eV_0\) is the maximum kinetic energy of emitted electrons, where \(e\) is the electron charge \((1.6 \times 10^{-19}\ \text{C})\), and \(V_0\) is the stopping potential.
• \(W\) is the work function of the metallic surface.

For the first light with wavelength \(\lambda\) and stopping potential \(V_0 = 3.2\ \text{V}\), the equation becomes:

\(\frac{hc}{\lambda} = e \times 3.2 + W \quad \text{(1)}\)

For the second light with wavelength \(2\lambda\) and stopping potential \(V'_0 = 0.7\ \text{V}\), the equation becomes:

\(\frac{hc}{2\lambda} = e \times 0.7 + W \quad \text{(2)}\)

Subtract equation (2) from equation (1):

\(\left(\frac{hc}{\lambda} - \frac{hc}{2\lambda}\right) = 3.2e - 0.7e\)

Simplifying, we find:

\(\frac{hc}{2\lambda} = 2.5e\)

Solving for \(\lambda\):

\(\lambda = \frac{hc}{2.5e}\)

Substitute the values of constants:

\(\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 1.6 \times 10^{-19}}\)

Evaluating gives:

\(\lambda \approx 2.5 \times 10^{-7}\ \text{m}\)

Hence, the wavelength of the first light is 2.5 x 10^-7 m.