The de-Broglie wavelength associated with a particle of mass `10^-6 kg` moving with a velocity of `10 ms^-1`, is

Options

A. `6.63 xx 10^-22 m`
B. `6.63 xx 10^-29 m`
C. `6.63 xx 10^-31 m`
D. `6.63 xx 10^-34 m`

Explanation

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

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

where:
- \(\lambda\) is the de-Broglie wavelength,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the particle,
- \(v\) is the velocity of the particle.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass (\(m\)) = \(10^{-6} \, \text{kg}\)
- Velocity (\(v\)) = \(10 \, \text{m/s}\)

2. **Substitute the values into the de-Broglie wavelength formula:**
\[ \lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34} \, \text{Js}}{(10^{-6} \, \text{kg})(10 \, \text{m/s})} \]

3. **Calculate the denominator:**
\[ mv = (10^{-6} \, \text{kg})(10 \, \text{m/s}) = 10^{-5} \, \text{kg m/s} \]

4. **Now substitute back into the equation:**
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{10^{-5} \, \text{kg m/s}} \]

5. **Perform the division:**
\[ \lambda = 6.626 \times 10^{-34} \div 10^{-5} = 6.626 \times 10^{-34 + 5} = 6.626 \times 10^{-29} \, \text{m} \]

6. **Final result:**
\[ \lambda \approx 6.626 \times 10^{-29} \, \text{m} \]

### Conclusion:
The de-Broglie wavelength associated with the particle is approximately \(6.626 \times 10^{-29} \, \text{m}\).