Question

The work function of a photosensitive metal is 0.5 eV. If photons of energy 1 eV and 2.5 eV are incident on this metal separately, then the ratio of maximum kinetic energies of ejected electrons will be

• A. 1 : 5
• B. 1 : 4
• C. 1 : 2
• D. 1 : 1

Hint:

In the photoelectric effect, the kinetic energy of the ejected electrons is the difference between the photon energy and the work function.

Answer 1

The Correct Option is C

Step 1: Understanding the photoelectric effect.
The energy of the ejected electrons is
given by the equation:
\[ K.E. = E_{\text{photon}} - \phi \] where \( K.E. \) is the kinetic energy of the ejected electron, \( E_{\text{photon}} \) is the energy of the incident photon, and \( \phi \) is the work function of the metal.
Step 2: Calculating kinetic energies for each photon.
- For the photon with energy 1 eV: \[ K.E. = 1 \, \text{eV} - 0.5 \, \text{eV} = 0.5 \, \text{eV} \] - For the photon with energy 2.5 eV: \[ K.E. = 2.5 \, \text{eV} - 0.5 \, \text{eV} = 2 \, \text{eV} \]
Step 3: Finding the ratio.
The ratio of kinetic energies is: \[ \frac{K.E._{1\,\text{eV}}}{K.E._{2.5\,\text{eV}}} = \frac{0.5}{2} = \frac{1}{4} \] Thus, the correct answer is
(C) 1 : 2.