Question

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

• A. 1 : 2
• B. 1:3
• C. 2 : 3
• D. 3 : 2

Answer 1

The Correct Option is C

To determine the ratio of maximum velocities of the photoelectrons emitted in the two cases, we need to apply the photoelectric effect equation. The photoelectric effect can be described by the equation:

K.E. = E - \phi

where:

• K.E. is the kinetic energy of the emitted photoelectron.
• E is the energy of the incident photon.
• \phi is the work function of the metal.

According to the kinetic energy equation, the maximum kinetic energy is also related to the velocity by:

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

where:

• m is the mass of the electron, and
• v is the velocity of the photoelectron.

Let's denote:

• E_1 = 5\phi for the first stream of photons.
• E_2 = 10\phi for the second stream of photons.

Then, the maximum kinetic energies for the two cases are:

K.E._1 = 5\phi - \phi = 4\phi

K.E._2 = 10\phi - \phi = 9\phi

Using the relation between kinetic energy and velocity:

\frac{1}{2} m v_1^2 = 4\phi

\frac{1}{2} m v_2^2 = 9\phi

To find the ratio of the maximum velocities, we equate:

\frac{v_1^2}{v_2^2} = \frac{4\phi}{9\phi}

\frac{v_1^2}{v_2^2} = \frac{4}{9}

Therefore:

\frac{v_1}{v_2} = \sqrt{\frac{4}{9}} = \frac{2}{3}

Thus, the ratio of the maximum velocities of the photoelectrons in the two cases is 2:3.