If the Planck's constant h=6.6×`10^(-34)` Js, the de Broglie wave length of a particle having momentum of 1.1×`10^(-24) kg m s^(-1)` will be

Options

A. 2`A^0`
B. 4`A^0`
C. 6`A^0`
D. 8`A^0`

Explanation

To find the de Broglie wavelength of a particle given its momentum, we can use the 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 \(p = 1.1 \times 10^{-24} \, \text{kg m s}^{-1}\)

### Step 2: Substitute the values into the formula
Using the values provided, we substitute them into the de Broglie wavelength formula:

\[ \lambda = \frac{6.6 \times 10^{-34} \, \text{Js}}{1.1 \times 10^{-24} \, \text{kg m s}^{-1}} \]

### Step 3: Perform the calculation
Now, we perform the division:

\[ \lambda = \frac{6.6}{1.1} \times \frac{10^{-34}}{10^{-24}} = 6 \times 10^{-10} \, \text{m} \]

### Step 4: Convert to Angstroms
Since \(1 \, \text{angstrom} = 10^{-10} \, \text{m}\), we can convert the wavelength from meters to angstroms:

\[ \lambda = 6 \times 10^{-10} \, \text{m} = 6 \, \text{angstrom} \]

### Final Answer
The de Broglie wavelength of the particle is \(6 \, \text{angstrom}\).

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