The de-Broglie wavelength of a particle having a momentum of `2xx10^(-28) kg m//s` is

Options

A. `3.3xx10^(-5)` m
B. `6.6xx10^(-6) m`
C. `3.3xx10^(-6) m`
D. `1.65xx10^(-6) m`

Explanation

To find the de-Broglie wavelength of a particle with a given momentum, we can use the de-Broglie wavelength formula:

\[ \lambda = \frac{h}{p} \]

where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is the Planck constant, approximately \(6.67 \times 10^{-34} \, \text{J s}\),
- \(p\) is the momentum of the particle.

### Step-by-Step Solution:

1. **Identify the given values**:
- Momentum \(p = 2 \times 10^{-28} \, \text{kg m/s}\)
- Planck constant \(h = 6.67 \times 10^{-34} \, \text{J s}\)

2. **Substitute the values into the de-Broglie wavelength formula**:
\[ \lambda = \frac{h}{p} = \frac{6.67 \times 10^{-34} \, \text{J s}}{2 \times 10^{-28} \, \text{kg m/s}} \]

3. **Calculate the wavelength**:
- First, calculate the denominator:
\[ 2 \times 10^{-28} = 2.0 \times 10^{-28} \]
- Now perform the division:
\[ \lambda = \frac{6.67 \times 10^{-34}}{2.0 \times 10^{-28}} = 3.335 \times 10^{-6} \, \text{m} \]

4. **Round the answer**:
- Rounding \(3.335 \times 10^{-6}\) gives approximately:
\[ \lambda \approx 3.3 \times 10^{-6} \, \text{m} \]

5. **Final Answer**:
- The de-Broglie wavelength of the particle is:
\[ \lambda \approx 3.3 \times 10^{-6} \, \text{m} \]