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: Apply the photoelectric equation. The photoelectric equation states: \[ K_{\max} = h\nu - \phi, \] where: - \( K_{\max} \) is the maximum kinetic energy of the emitted electrons, - \( h\nu \) is the photon energy, - \( \phi \) is the work function of the metal. Step 2: Determine the photon energy for initial wavelength \( \lambda \). Since the initial kinetic energy is 2 eV, we write: \[ h \nu = K_{\max} + \phi = 2 + 1 = 3 \text{ eV}. \] Since \( h \nu = \frac{hc}{\lambda} \), we set: \[ \frac{hc}{\lambda} = 3 \text{ eV}. \] Step 3: Compute the new kinetic energy for \( \lambda/2 \). The photon energy for \( \lambda/2 \) is: \[ h \nu' = \frac{hc}{\lambda/2} = 2 \times \frac{hc}{\lambda} = 2 \times 3 = 6 \text{ eV}. \] Thus, the new kinetic energy is: \[ K_{\max}' = 6 - 1 = 5 \text{ eV}. \] Thus, the answer is \( \boxed{5} \).

Answer 2

Step 1: Use Einstein’s photoelectric equation. 

\[ K_{\max} = h\nu - \phi \] where \( K_{\max} \) = maximum kinetic energy of the emitted electrons, \( \nu \) = frequency of incident light, \( \phi \) = work function of the metal.

 

Step 2: For wavelength \( \lambda \):

\[ K_1 = h\nu_1 - \phi = 2\,\text{eV} \] and given \( \phi = 1\,\text{eV} \), \[ h\nu_1 = K_1 + \phi = 2 + 1 = 3\,\text{eV}. \]

Step 3: For wavelength \( \frac{\lambda}{2} \):

Frequency doubles, since \( \nu \propto \frac{1}{\lambda} \): \[ \nu_2 = 2\nu_1. \] Hence the new photon energy: \[ h\nu_2 = 2h\nu_1 = 2 \times 3 = 6\,\text{eV}. \]

Step 4: Compute new kinetic energy.

\[ K_2 = h\nu_2 - \phi = 6 - 1 = 5\,\text{eV}. \] 

---

Final Answer:

\[ \boxed{K_{\max} = 5\,\text{eV}} \]