If the Planck's constant h=6.6×`10^(-34)` Js, the de Broglie wave length of a particle having momentum of 5.5 .×`10^(-24) kg m s^(-1)` will be

Options

A. 1.2`A^0`
B. 2.2`A^0`
C. 3.2`A^0`
D. 4.2`A^0`

Explanation

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

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

where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is Planck's constant,
- \(p\) is the momentum of the particle.

### Step-by-step Solution:

1. **Identify the values given in the problem:**
- Planck's constant \(h = 6.6 \times 10^{-34} \, \text{Js}\)
- Momentum \(p = 5.5 \times 10^{-24} \, \text{kg m s}^{-1}\)

2. **Substitute the values into the de Broglie wavelength formula:**
\[ \lambda = \frac{h}{p} = \frac{6.6 \times 10^{-34} \, \text{Js}}{5.5 \times 10^{-24} \, \text{kg m s}^{-1}} \]

3. **Perform the calculation:**
- First, calculate the division:
\[ \lambda = \frac{6.6}{5.5} \times \frac{10^{-34}}{10^{-24}} = 1.2 \times 10^{-10} \, \text{m} \]

4. **Convert the wavelength into angstroms:**
- Since \(1 \, \text{angstrom} = 10^{-10} \, \text{m}\), we can express the wavelength in angstroms:
\[ \lambda = 1.2 \, \text{angstrom} \]

5. **Final Answer:**
The de Broglie wavelength of the particle is \(1.2 \, \text{angstrom}\).