Question

Radiations of two frequencies are incident on a metal surface of work function 2.0 eV one by one. The energies of their photons are 2.5 eV and 4.5 eV respectively. Find the ratio of the maximum speed of the electrons emitted in the two cases.

Hint:

The kinetic energy of the emitted electron depends on the energy of the incoming photon and the work function of the material.

Answer 1

Step 1: Apply the photoelectric equation. The maximum kinetic energy (KE) of an emitted electron is determined by the photoelectric equation: \[ K.E. = hu - \phi \] In this equation, \( hu \) represents the energy of the incident photon, and \( \phi \) is the work function of the metal.

Step 2: Calculate kinetic energy for each scenario. For the first scenario, where the photon energy is 2.5 eV: \[ K.E_1 = 2.5 \, \text{eV} - 2.0 \, \text{eV} = 0.5 \, \text{eV} \] For the second scenario, with a photon energy of 4.5 eV: \[ K.E_2 = 4.5 \, \text{eV} - 2.0 \, \text{eV} = 2.5 \, \text{eV} \]

Step 3: Determine the maximum speed of emitted electrons. The relationship between kinetic energy and speed is given by: \[ K.E. = \frac{1}{2} m v^2 \] where \( m \) is the electron's mass and \( v \) is its speed. Therefore, the ratio of the maximum speeds (\( v_2/v_1 \)) can be calculated as: \[ \frac{v_2}{v_1} = \sqrt{\frac{K.E_2}{K.E_1}} = \sqrt{\frac{2.5}{0.5}} = \sqrt{5} \approx 2.24 \] The ratio of the maximum speeds of the electrons is approximately 2.24.