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

Options

A. 1`A^0`
B. 2`A^0`
C. 3`A^0`
D. 4`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 1: Identify the given values
- Planck's constant \(h = 6.6 \times 10^{-34} \, \text{Js}\)
- Momentum \(p = 6.6 \times 10^{-24} \, \text{kg m/s}\)

### Step 2: Substitute the values into the formula
Now we substitute the values of \(h\) and \(p\) into the de Broglie wavelength formula:

\[ \lambda = \frac{6.6 \times 10^{-34} \, \text{Js}}{6.6 \times 10^{-24} \, \text{kg m/s}} \]

### Step 3: Simplify the equation
When we divide the two values, we can simplify:

\[ \lambda = \frac{6.6}{6.6} \times \frac{10^{-34}}{10^{-24}} = 1 \times 10^{-10} \, \text{m} \]

### Step 4: Convert to Angstroms
Since \(1 \, \text{Angstrom} = 10^{-10} \, \text{m}\), we can express the wavelength in Angstroms:

\[ \lambda = 1 \, \text{Angstrom} \]

### Final Answer
Thus, the de Broglie wavelength of the particle is:

\[ \lambda = 1 \, \text{Angstrom} \]