When photons of energy hv are incident on the surface of photosensitive material of work function `hv_0`, then

Options

A. the kinetic energy of all emitted electrons is `hv_0`
B. the kinetic energy of all emitted electrons is `h(v-v_0)`
C. the kinetic energy of all fastest emitted electrons is `h(v-v_0)`
D. the kinetic energy of all emitted electrons is hv

Answer

To solve the problem regarding the photoelectric effect, we will use Einstein's photoelectric equation and analyze the options provided in the question.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect occurs when light (photons) hits a material and causes the emission of electrons. The energy of the incoming photons is given by \( E = h\nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency of the light.

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

3. **Einstein's Photoelectric Equation**:
The maximum kinetic energy \( K \) of the emitted electrons can be expressed as:
\[ K = h\nu - h\nu_0 \]
This equation indicates that the kinetic energy of the emitted electrons is equal to the energy of the incoming photons minus the work function of the material.

4. **Analyzing the Options**:
- **Option 1**: "The kinetic energy of all emitted electrons is \( h\nu_0 \)".
- This is incorrect because the kinetic energy cannot equal the work function; it must be the difference between the photon energy and the work function.

- **Option 2**: "The kinetic energy of all emitted electrons is \( h\nu - h\nu_0 \)".
- This is misleading. While the expression \( h\nu - h\nu_0 \) is correct for the maximum kinetic energy, it does not apply to all emitted electrons since different electrons may lose varying amounts of energy due to collisions.

- **Option 3**: "The kinetic energy of all fastest electrons is \( h\nu - h\nu_0 \)".
- This is correct. The fastest electrons (those with maximum kinetic energy) will have kinetic energy equal to the photon energy minus the work function.

- **Option 4**: "The kinetic energy of all emitted electrons is \( h\nu \)".
- This is incorrect because it implies that all the energy of the photon goes into kinetic energy, ignoring the work function.

5. **Conclusion**:
The correct answer is **Option 3**: "The kinetic energy of all fastest electrons is \( h\nu - h\nu_0 \)".