Question

A photosensitive metallic surface has work function, hv₀. If photons of energy 2hv₀ fall on this surface, the electrons come out with a maximum velocity of 4x10⁶ m/s. When the photon energy is increased to 5hv₀, then maximum velocity of photoelectrons will be:

• A. 16x10⁶ m/s
• B. 8x10⁷ m/s
• C. 4x10⁵ m/s
• D. 8x10⁶ m/s

Answer 1

The Correct Option is D

To solve this problem, we will use the photoelectric effect equation. The photoelectric effect describes the emission of electrons from a metal when it absorbs electromagnetic radiation. The maximum kinetic energy of the emitted electrons can be described by the equation:

K.E_{\text{max}} = h\nu - \phi

Where:

• K.E_{\text{max}} is the maximum kinetic energy of the emitted photoelectrons.
• h\nu is the energy of the incident photons.
• \phi is the work function of the metal surface.

Given that:

• For the initial condition: {h\nu = 2hv_0}, where v_0 is the threshold frequency.
• The maximum velocity of electrons v = 4 \times 10^6 \, \text{m/s}.

The kinetic energy of the electrons is given by:

K.E_{\text{max}} = \frac{1}{2}mv^2

Let's calculate the kinetic energy for the initial condition. Substitute v = 4 \times 10^6 \, \text{m/s}:

K.E_{\text{max}} = \frac{1}{2} m (4 \times 10^6)^2

This kinetic energy equals the excess energy from the photons:

2hv_0 - hv_0 = \frac{1}{2}m (4 \times 10^6)^2

Now, increase the photon energy to 5hv_0 and calculate the new kinetic energy:

K.E'_{\text{max}} = 5hv_0 - hv_0 = 4hv_0

Relate this to the new velocity v':

\frac{1}{2}m {v'}^2 = 4\times \frac{1}{2}m (4 \times 10^6)^2

Cancel \frac{1}{2}m from both sides and solve for {v'}^2:

{v'}^2 = 4(4 \times 10^6)^2

Therefore, the new velocity v' is:

v' = \sqrt{16} \times 10^6 = 4 \times 10^6 \times 2 = 8 \times 10^6 \, \text{m/s}

Thus, the maximum velocity of the photoelectrons when the photon energy is increased to 5hv_0 is 8 \times 10^6 \, \text{m/s}.

The correct answer is: 8 x 10⁶ m/s