When the light of frequency `2v_0` where `v_0` is threshold frequency is incident on a metal plate . the maximum velocity of electrons emitted is `v_1` when the frequency of the incident radiation is increased to `5v_0` the maximum velocity of electrons emitted from the same plate is `v_2` The ratio of `v_1 " to" v_2`

Options

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

Explanation

To solve the problem, we need to use the concept of the photoelectric effect, which relates the frequency of incident light to the kinetic energy of emitted electrons. Here's a step-by-step solution:

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**: The maximum kinetic energy (KE) of the emitted electrons can be expressed as:
\[ KE = h\nu - h\nu_0 \]
where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \nu_0 \) is the threshold frequency.

2. **Set Up the Equations for the Two Frequencies**:
- For the first case where the frequency is \( 2\nu_0 \):
\[ KE_1 = h(2\nu_0) - h\nu_0 = h\nu_0 \]
The maximum kinetic energy can also be expressed in terms of the velocity \( v_1 \):
\[ KE_1 = \frac{1}{2}mv_1^2 \]
Thus, we have:
\[ \frac{1}{2}mv_1^2 = h\nu_0 \quad \text{(Equation 1)} \]

- For the second case where the frequency is \( 5\nu_0 \):
\[ KE_2 = h(5\nu_0) - h\nu_0 = 4h\nu_0 \]
Similarly, the maximum kinetic energy can be expressed as:
\[ KE_2 = \frac{1}{2}mv_2^2 \]
Thus, we have:
\[ \frac{1}{2}mv_2^2 = 4h\nu_0 \quad \text{(Equation 2)} \]

3. **Divide the Two Equations**:
To find the ratio of \( v_1 \) to \( v_2 \), we divide Equation 1 by Equation 2:
\[ \frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{h\nu_0}{4h\nu_0} \]
The \( \frac{1}{2}m \) and \( h\nu_0 \) terms cancel out:
\[ \frac{v_1^2}{v_2^2} = \frac{1}{4} \]

4. **Take the Square Root**:
Taking the square root of both sides gives:
\[ \frac{v_1}{v_2} = \frac{1}{2} \]

5. **Final Ratio**:
Therefore, the ratio of the maximum velocities of the emitted electrons is:
\[ v_1 : v_2 = 1 : 2 \]