If the Planck's constant h=6.6×`10^(-34)` Js, the de Broglie wave length of a particle having momentum of 3.3×`10^(-24) kg m s^(-1)` will be ...

Options

A. 2`A^0`
B. 0.002`A^0`
C. 0.5`A^0`
D. 5`A^0`

Explanation

To find the de Broglie wavelength of a particle given its momentum and Planck's constant, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula**: The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Substitute the Given Values**:
- Given Planck's constant \( h = 6.6 \times 10^{-34} \) Js
- Given momentum \( p = 3.3 \times 10^{-24} \) kg m/s

Substitute these values into the formula:
\[ \lambda = \frac{6.6 \times 10^{-34} \text{ Js}}{3.3 \times 10^{-24} \text{ kg m/s}} \]

3. **Perform the Calculation**:
- Calculate the division:
\[ \lambda = \frac{6.6}{3.3} \times \frac{10^{-34}}{10^{-24}} = 2 \times 10^{-10} \text{ m} \]

4. **Convert to Angstroms**:
- Since \( 1 \text{ Å} = 10^{-10} \text{ m} \), we can convert the wavelength from meters to angstroms:
\[ \lambda = 2 \times 10^{-10} \text{ m} = 2 \text{ Å} \]

5. **Final Result**:
- The de Broglie wavelength of the particle is:
\[ \lambda = 2 \text{ Å} \]