Threshold frequency of metal is `f_(0)`. When light of frequency `v = 2f_(0)` is incident on the metal plate, velocity of electron emitted in `V_(1)`. When a plate frequency of incident radiation is `5f_(0), V_(2)` is velocity of emitted electron, then `V_(1):V_(2)` is

Options

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

Explanation

To solve the problem, we need to use the photoelectric effect equation, which relates the kinetic energy of the emitted electrons to the frequency of the incident light. The equation is given by:

\[ KE = h\nu - h\nu_0 \]

Where:
- \( KE \) is the kinetic energy of the emitted electron.
- \( h \) is Planck's constant.
- \( \nu \) is the frequency of the incident light.
- \( \nu_0 \) is the threshold frequency of the metal.

### Step 1: Calculate Kinetic Energy for \( V_1 \)
When the frequency of the incident light is \( \nu = 2\nu_0 \):

Using the equation for kinetic energy:
\[ KE_1 = h(2\nu_0) - h\nu_0 \]
\[ KE_1 = h(2\nu_0 - \nu_0) \]
\[ KE_1 = h\nu_0 \]

The kinetic energy can also be expressed in terms of the velocity of the emitted electrons:
\[ KE_1 = \frac{1}{2} mv_1^2 \]

Setting these equal gives:
\[ \frac{1}{2} mv_1^2 = h\nu_0 \]

### Step 2: Calculate Kinetic Energy for \( V_2 \)
When the frequency of the incident light is \( \nu = 5\nu_0 \):

Using the equation for kinetic energy:
\[ KE_2 = h(5\nu_0) - h\nu_0 \]
\[ KE_2 = h(5\nu_0 - \nu_0) \]
\[ KE_2 = 4h\nu_0 \]

Again, expressing this in terms of velocity:
\[ KE_2 = \frac{1}{2} mv_2^2 \]

Setting these equal gives:
\[ \frac{1}{2} mv_2^2 = 4h\nu_0 \]

### Step 3: Relate \( V_1 \) and \( V_2 \)
Now we have two equations:
1. \( \frac{1}{2} mv_1^2 = h\nu_0 \)
2. \( \frac{1}{2} mv_2^2 = 4h\nu_0 \)

From the first equation:
\[ v_1^2 = \frac{2h\nu_0}{m} \]

From the second equation:
\[ v_2^2 = \frac{8h\nu_0}{m} \]

### Step 4: Find the Ratio \( V_1 : V_2 \)
To find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1^2}{v_2^2} = \frac{\frac{2h\nu_0}{m}}{\frac{8h\nu_0}{m}} = \frac{2}{8} = \frac{1}{4} \]

Taking the square root to find the ratio of velocities:
\[ \frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

Thus, the ratio \( V_1 : V_2 \) is:
\[ V_1 : V_2 = 1 : 2 \]

### Final Answer
The answer to the question is \( 1 : 2 \).