The momentum of a particle associated with de-Broglie's wavelength of `6 Å` is

Options

A. `1.1 xx 10^(-24) kg ms^(-1)`
B. `1.1 xx 10^(34) kg ms^(-1)`
C. `39.6 xx 10^(-34) kg ms^(-1)`
D. `39.6 xx 10^(-24) kg ms^(-1)`

Explanation

To find the momentum of a particle associated with a de-Broglie wavelength of 6 Å (angstroms), we can use the de-Broglie wavelength formula:

### Step-by-Step Solution:

1. **Understand the de-Broglie Wavelength Formula**:
The de-Broglie wavelength (\( \lambda \)) is related to the momentum (p) of a particle by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is the Planck constant.

2. **Rearrange the Formula to Solve for Momentum**:
We can rearrange the formula to express momentum in terms of wavelength:
\[ p = \frac{h}{\lambda} \]

3. **Substitute the Values**:
- The Planck constant \( h \) is approximately \( 6.63 \times 10^{-34} \, \text{kg m}^2/\text{s} \).
- The given wavelength \( \lambda \) is \( 6 \, \text{Å} \). We need to convert this into meters:
\[ 6 \, \text{Å} = 6 \times 10^{-10} \, \text{m} \]

4. **Calculate the Momentum**:
Now, substituting the values into the momentum formula:
\[ p = \frac{6.63 \times 10^{-34} \, \text{kg m}^2/\text{s}}{6 \times 10^{-10} \, \text{m}} \]
Performing the calculation:
\[ p = \frac{6.63 \times 10^{-34}}{6 \times 10^{-10}} = 1.105 \times 10^{-24} \, \text{kg m/s} \]

5. **Final Result**:
The momentum of the particle associated with a de-Broglie wavelength of 6 Å is approximately:
\[ p \approx 1.1 \times 10^{-24} \, \text{kg m/s} \]