Question

When the light of frequency 2v₀ (where v₀ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v₁ . When the frequency of the incident radiation is increased to 5v₀ , the maximum velocity of electrons emitted from the same plate is v₂ . The ratio of v₁ to v₂ is

• A. 1:2
• B. 4:1
• C. 1:4
• D. 2:1

Answer 1

The Correct Option is A

To solve this problem, we use the photoelectric effect equation which is derived from Einstein's photoelectric equation:

K.E. = h\nu - \phi

where:

• K.E. is the kinetic energy of the emitted electron.
• h is Planck's constant.
• \nu is the frequency of the incident light.
• \phi is the work function of the metal, equal to h\nu_0 (where \nu_0 is the threshold frequency).

Given conditions:

• Initial frequency: \nu = 2\nu_0
• Increased frequency: \nu = 5\nu_0

For the initial condition where \nu = 2\nu_0, the kinetic energy is given by:

K.E._1 = h(2\nu_0) - h\nu_0 = h\nu_0

The kinetic energy is also given by:

\frac{1}{2} m v_1^2 = h\nu_0

(1)

For the second condition where \nu = 5\nu_0, the kinetic energy is given by:

K.E._2 = h(5\nu_0) - h\nu_0 = 4h\nu_0

So, the equation becomes:

\frac{1}{2} m v_2^2 = 4h\nu_0

(2)

To find the ratio \frac{v_1}{v_2}, we divide equation (1) by equation (2):

\frac{\frac{1}{2} m v_1^2}{\frac{1}{2} m v_2^2} = \frac{h\nu_0}{4h\nu_0}

\Rightarrow \frac{v_1^2}{v_2^2} = \frac{1}{4}

Taking square roots on both sides, we get:

\Rightarrow \frac{v_1}{v_2} = \frac{1}{2}

Hence, the ratio of v_1 to v_2 is 1:2.