The energy of a photon of frequency v is E = hv and the momentum of a photon of wavelength `lambda` is p = `h/lambda` From this statement one may conclude that the wave velocity of light is equal to :

Options

A. `3 xx 10^(8) ms^(-1)`
B. `(E)/(P)`
C. Ep
D. `((E)/(P))^(2)`

Explanation

To find the wave velocity of light based on the given relationships for the energy and momentum of a photon, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Relationships**:
- The energy \( E \) of a photon is given by the equation:
\[ E = h \nu \]
- The momentum \( p \) of a photon is given by the equation:
\[ p = \frac{h}{\lambda} \]

2. **Recall the Wave Velocity Formula**:
- The wave velocity \( v \) of any wave can be expressed as:
\[ v = \lambda \nu \]
- Here, \( \lambda \) is the wavelength and \( \nu \) is the frequency of the wave.

3. **Substitute Frequency in Terms of Velocity and Wavelength**:
- From the wave velocity formula, we can express frequency \( \nu \) as:
\[ \nu = \frac{v}{\lambda} \]

4. **Substitute Frequency into the Energy Equation**:
- Now, substituting \( \nu \) into the energy equation:
\[ E = h \nu = h \left(\frac{v}{\lambda}\right) \]
- This simplifies to:
\[ E = \frac{hv}{\lambda} \]

5. **Relate Energy and Momentum**:
- From the momentum equation, we know:
\[ p = \frac{h}{\lambda} \]
- Rearranging this gives:
\[ h = p \lambda \]

6. **Substitute \( h \) into the Energy Equation**:
- Now substitute \( h \) into the energy equation:
\[ E = \frac{(p \lambda)v}{\lambda} \]
- The \( \lambda \) cancels out:
\[ E = pv \]

7. **Solve for Wave Velocity**:
- Rearranging the equation \( E = pv \) gives:
\[ v = \frac{E}{p} \]

8. **Conclusion**:
- Therefore, the wave velocity of light is given by:
\[ v = \frac{E}{p} \]

### Final Answer:
The wave velocity of light is equal to \( \frac{E}{p} \).

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