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: Using the photoelectric equation. The maximum kinetic energy of the emitted electron is given by: \[ K.E. = h\nu - \phi \] Where \( h\nu \) is the energy of the photon and \( \phi \) is the work function of the metal.

Step 2: Kinetic energy for each case. For the first case with photon energy 2.5 eV: \[ K.E_1 = 2.5 \, \text{eV} - 2.0 \, \text{eV} = 0.5 \, \text{eV} \] For the second case with photon energy 4.5 eV: \[ K.E_2 = 4.5 \, \text{eV} - 2.0 \, \text{eV} = 2.5 \, \text{eV} \]

Step 3: Maximum speed of emitted electrons. The kinetic energy of the electron is related to its speed by: \[ K.E. = \frac{1}{2} m v^2 \] Where \( m \) is the mass of the electron and \( v \) is its speed. Thus, the ratio of the maximum speeds is: \[ \frac{v_2}{v_1} = \sqrt{\frac{K.E_2}{K.E_1}} = \sqrt{\frac{2.5}{0.5}} = \sqrt{5} \approx 2.24 \] Thus, the ratio of the maximum speeds of the electrons is approximately 2.24.