Question

The radiations of energies 1eV and 2.5eV are incident on a metal surface having work function 0.5 eV. The ratio of the maximum velocities of the emitted photo-electrons is:

• A. 1:1
• B. 1:2
• C. 1:3
• D. 1:4

Hint:

The maximum velocity of a photoelectron is proportional to the square root of its kinetic energy.

Answer 1

The Correct Option is B

Step 1: Kinetic Energy of Photo-electrons.
The kinetic energy of a photo-electron is given by: \[ K.E. = E_{\text{photon}} - \phi \] where \( \phi \) is the work function of the metal. For the first radiation with energy 1 eV: \[ K.E_1 = 1 - 0.5 = 0.5 \, \text{eV} \] For the second radiation with energy 2.5 eV: \[ K.E_2 = 2.5 - 0.5 = 2 \, \text{eV} \] The maximum velocity of the emitted photo-electrons is related to the kinetic energy by: \[ K.E = \frac{1}{2} mv^2 \] where \( m \) is the mass of the electron and \( v \) is its velocity. Thus, the ratio of the velocities is: \[ \frac{v_2}{v_1} = \sqrt{\frac{K.E_2}{K.E_1}} = \sqrt{\frac{2}{0.5}} = \sqrt{4} = 2 \] Step 2: Final Answer.
Thus, the ratio of the velocities is 1:2.