The Threshold energy is given as `E_(0)` and radiation of energy E falls on metal then ` K.E` is given as

Options

A. `(E-E_0)/(2)`
B. `E-E_0`
C. `E_0 -E`
D. `(E )/(E _0)`

Answer

To solve the problem, we need to understand the relationship between the energy of the incident radiation, the threshold energy (also known as the work function), and the kinetic energy of the emitted electrons.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- The threshold energy (work function) is denoted as \( E_0 \).
- The energy of the incident radiation is denoted as \( E \).

2. **Understand the Concept**:
- When radiation of energy \( E \) falls on a metal, it can eject electrons from the surface of the metal if the energy is greater than the threshold energy \( E_0 \).
- The excess energy after overcoming the threshold energy is converted into kinetic energy of the emitted electrons.

3. **Write the Energy Equation**:
- The relationship between the incident energy, the work function, and the kinetic energy of the emitted electrons can be expressed as:
\[ E = E_0 + K.E. \]
- Here, \( K.E. \) is the maximum kinetic energy of the emitted electrons.

4. **Rearrange the Equation**:
- To find the kinetic energy, rearrange the equation:
\[ K.E. = E - E_0. \]

5. **Final Expression**:
- Thus, the kinetic energy of the emitted electrons when radiation of energy \( E \) falls on the metal is given by:
\[ K.E. = E - E_0. \]

### Conclusion:
The kinetic energy of the emitted electrons is \( K.E. = E - E_0 \).