Question

A light source of wavelength \( \lambda \) illuminates a metal surface, and electrons are ejected with a maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength \( \frac{\lambda}{2} \), then the maximum kinetic energy of ejected electrons will be (The work function of the metal is 1 eV).

• A. 6 eV
• B. 5 eV
• C. 2 eV
• D. 3 eV

Hint:

In the photoelectric effect: - The photon energy is inversely proportional to the wavelength. - Doubling the frequency (or halving the wavelength) doubles the photon energy. - The kinetic energy of emitted electrons is given by \( K_{\max} = h\nu - \phi \).

Answer 1

The Correct Option is B

Step 1: State the photoelectric equation. The photoelectric equation is given by: \[ K_{\max} = hu - \phi, \] where $ K_{\max} $ represents the maximum kinetic energy of the ejected electrons, $ hu $ is the energy of the incident photon, and $ \phi $ is the work function of the metallic surface. Step 2: Calculate the photon energy for the initial wavelength $ \lambda $. Given that the initial kinetic energy is 2 eV, we can express the photon energy as: \[ h u = K_{\max} + \phi = 2 + 1 = 3 \text{ eV}. \] Since the photon energy is related to wavelength by $ h u = \frac{hc}{\lambda} $, we have: \[ \frac{hc}{\lambda} = 3 \text{ eV}. \] Step 3: Compute the new kinetic energy when the wavelength is halved ($ \lambda/2 $). The photon energy corresponding to the new wavelength $ \lambda/2 $ is: \[ h u' = \frac{hc}{\lambda/2} = 2 \times \frac{hc}{\lambda} = 2 \times 3 = 6 \text{ eV}. \] The new maximum kinetic energy is therefore: \[ K_{\max}' = 6 - 1 = 5 \text{ eV}. \] The result is $ \boxed{5} $.