When the photons of energy hv fall on a photosensitive metallic surface of work function `hv_(0)`, electrons are emitted are from the surface. The most energetic electron coming out of the surfece have kinetic energy equal to

Options

A. hv
B. `hv_(0)`
C. `hv+hv_(0)`
D. `hv-hv_(0)`

Explanation

To find the kinetic energy of the most energetic electron emitted from a photosensitive metallic surface when photons of energy \( h\nu \) fall on it, we can use the principle of energy conservation. Here are the steps to derive the solution:

### Step-by-Step Solution:

1. **Understand the Concept**: When a photon with energy \( h\nu \) strikes a photosensitive surface, it can transfer its energy to an electron in the metal. If the energy of the photon is greater than the work function of the metal \( h\nu_0 \), the electron can be emitted.

2. **Define Work Function**: The work function \( h\nu_0 \) is the minimum energy required to remove an electron from the surface of the metal.

3. **Energy Conservation**: The energy of the incoming photon is used to overcome the work function and any remaining energy is converted into the kinetic energy of the emitted electron.

4. **Write the Equation**: The maximum kinetic energy \( K.E_{max} \) of the emitted electron can be expressed as:
\[ K.E_{max} = \text{Energy of photon} - \text{Work function} \]
Substituting the known values:
\[ K.E_{max} = h\nu - h\nu_0 \]

5. **Final Expression**: Thus, the maximum kinetic energy of the most energetic electron emitted from the surface is given by:
\[ K.E_{max} = h\nu - h\nu_0 \]

### Final Answer:
The most energetic electron coming out of the surface has kinetic energy equal to \( K.E_{max} = h\nu - h\nu_0 \).

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