Light of frequency `v` is incident on a substance of threshold frequency `v(v lt v)`. The energy of the emitted photo-electron will be

Options

A. `h(v-v_(0))`
B. `h//v`
C. `he(v-v_(0))`
D. `h//v_(0)`

Answer

To find the energy of the emitted photo-electron when light of frequency \( v \) is incident on a substance with a threshold frequency \( v_0 \) (where \( v < v_0 \)), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**: The photoelectric effect states that when light of sufficient frequency shines on a material, it can eject electrons from that material. The energy of the incident photons must be greater than the work function of the material for electrons to be emitted.

2. **Identify Key Variables**:
- Let \( v \) be the frequency of the incident light.
- Let \( v_0 \) be the threshold frequency of the material.
- The energy of a photon is given by \( E = h \cdot v \), where \( h \) is Planck's constant.

3. **Determine the Work Function**: The work function \( \phi \) of the material is related to the threshold frequency by the equation:
\[ \phi = h \cdot v_0 \]

4. **Calculate the Energy of the Incident Photon**: The energy of the incident photon can be calculated as:
\[ E = h \cdot v \]

5. **Apply the Photoelectric Equation**: According to the photoelectric effect, the maximum kinetic energy \( K_{max} \) of the emitted photo-electron is given by:
\[ K_{max} = E - \phi \]
Substituting the expressions for \( E \) and \( \phi \):
\[ K_{max} = h \cdot v - h \cdot v_0 \]

6. **Factor Out Planck's Constant**: We can factor out \( h \) from the equation:
\[ K_{max} = h (v - v_0) \]

7. **Conclusion**: Since \( v < v_0 \), the energy of the emitted photo-electron will be negative, indicating that no photo-electron is emitted. Therefore, the energy of the emitted photo-electron is zero, as the incident light does not have enough energy to overcome the work function.

### Final Answer:
The energy of the emitted photo-electron will be zero since \( v < v_0 \).