An electtron and a photon have same wavelength . If `p` is the moment of electron and `E` the energy of photons, the magnitude of `p//E` in S I unit is

Options

A. `3.0xx10^(8)`
B. `3.33xx10^(-9)`
C. `9.1xx10^(-31)`
D. `6.64xx10^(-34)`

Answer

To solve the problem, we need to find the ratio of the momentum of the electron (p) to the energy of the photon (E) when both have the same wavelength (λ).

### Step-by-Step Solution:

1. **Understanding the Relationships**:
- The momentum (p) of an electron can be expressed using the de Broglie wavelength formula:
\[ p = \frac{h}{\lambda} \]
where \( h \) is Planck's constant.

- The energy (E) of a photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( c \) is the speed of light.

2. **Finding the Ratio \( \frac{p}{E} \)**:
- We need to find the ratio \( \frac{p}{E} \):
\[ \frac{p}{E} = \frac{\frac{h}{\lambda}}{\frac{hc}{\lambda}} \]

3. **Simplifying the Ratio**:
- When we simplify the ratio, we can cancel out \( \lambda \) and \( h \):
\[ \frac{p}{E} = \frac{h}{\lambda} \cdot \frac{\lambda}{hc} = \frac{1}{c} \]

4. **Substituting the Value of \( c \)**:
- The speed of light \( c \) is approximately \( 3 \times 10^8 \) m/s. Therefore:
\[ \frac{p}{E} = \frac{1}{3 \times 10^8} \]

5. **Calculating the Final Value**:
- Now, calculating \( \frac{1}{3 \times 10^8} \):
\[ \frac{1}{3 \times 10^8} \approx 3.33 \times 10^{-9} \]

### Conclusion:
The magnitude of \( \frac{p}{E} \) in SI units is \( 3.33 \times 10^{-9} \).

### Final Answer:
**Option B: \( 3.33 \times 10^{-9} \)**