Calculate the momentum of a particle which has a de Broglie wavelength of `2 Å,(h = 6.6 xx 10^(-34) kg m^(2) s^(-1))`

Explanation

To calculate the momentum of a particle with a de Broglie wavelength of 2 Å, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) is related to the momentum (p) of a particle by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

### Step 2: Rearrange the formula to find momentum
To find the momentum, we can rearrange the formula:
\[ p = \frac{h}{\lambda} \]

### Step 3: Convert the wavelength from angstroms to meters
Given that \( \lambda = 2 \, \text{Å} \), we need to convert this to meters.
1 Å (angstrom) = \( 10^{-10} \) meters, so:
\[ \lambda = 2 \, \text{Å} = 2 \times 10^{-10} \, \text{m} \]

### Step 4: Substitute the values into the momentum formula
Now, we can substitute the values into the momentum formula. We know:
- \( h = 6.626 \times 10^{-34} \, \text{kg m}^2 \text{s}^{-1} \)
- \( \lambda = 2 \times 10^{-10} \, \text{m} \)

Thus, we have:
\[ p = \frac{6.626 \times 10^{-34} \, \text{kg m}^2 \text{s}^{-1}}{2 \times 10^{-10} \, \text{m}} \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ p = \frac{6.626 \times 10^{-34}}{2 \times 10^{-10}} = 3.313 \times 10^{-24} \, \text{kg m/s} \]

### Step 6: Round to appropriate significant figures
Rounding to three significant figures, we get:
\[ p \approx 3.31 \times 10^{-24} \, \text{kg m/s} \]

### Final Answer
The momentum of the particle is approximately:
\[ p \approx 3.31 \times 10^{-24} \, \text{kg m/s} \]

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