Question

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be

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

Answer 1

The Correct Option is B

To determine the ratio of maximum speeds of electrons emitted when illuminated with two different photon energies, we can use the photoelectric effect, which is described by the equation:

K.E_{\text{max}} = E_{\text{photon}} - \phi,

where K.E_{\text{max}} is the maximum kinetic energy of the emitted electron, E_{\text{photon}} is the energy of the incident photon, and \phi is the work function of the metal. The relationship between kinetic energy and speed v of an electron is given by:

K.E = \frac{1}{2}mv^2.

Let's calculate the maximum kinetic energies for both photon energies:

1. For the photon energy of 3.8 eV:
   • The maximum kinetic energy is: K.E_{1} = 3.8 \, \text{eV} - 0.6 \, \text{eV} = 3.2 \, \text{eV}.
   • In joules, this is 3.2 \times 1.6 \times 10^{-19} \, \text{J}.
2. For the photon energy of 1.4 eV:
   • The maximum kinetic energy is: K.E_{2} = 1.4 \, \text{eV} - 0.6 \, \text{eV} = 0.8 \, \text{eV}.
   • In joules, this is 0.8 \times 1.6 \times 10^{-19} \, \text{J}.

The kinetic energy of the emitted electrons is related to their maximum speed v_{\text{max}} by the equation: K.E = \frac{1}{2}mv^2. Therefore, the speed is proportional to the square root of the kinetic energy:

v \propto \sqrt{\text{K.E}}.

Thus, the ratio of maximum speeds for the two frequencies is given by:

\frac{v_1}{v_2} = \sqrt{\frac{\text{K.E}_{1}}{\text{K.E}_{2}}} = \sqrt{\frac{3.2}{0.8}} = \sqrt{4} = 2.

Hence, the correct option is 2:1.