If the Planck's constant h=6.6×`10^(-34)` Js, the de Broglie wave length of a particle having momentum of 2.2×`10^(-24) kg m s^(-1)` will be

Options

A. 2`A^0`
B. 4`A^0`
C. 3`A^0`
D. 8`A^0`

Answer

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 values given in the problem
- Planck's constant \(h = 6.6 \times 10^{-34} \, \text{Js}\)
- Momentum \(p = 2.2 \times 10^{-24} \, \text{kg m/s}\)

### Step 2: Substitute the values into the de Broglie wavelength formula
\[ \lambda = \frac{6.6 \times 10^{-34} \, \text{Js}}{2.2 \times 10^{-24} \, \text{kg m/s}} \]

### Step 3: Perform the calculation
Calculating the right-hand side:
\[ \lambda = \frac{6.6}{2.2} \times \frac{10^{-34}}{10^{-24}} = 3 \times 10^{-10} \, \text{m} \]

### Step 4: Convert the wavelength to angstroms
Since \(1 \, \text{angstrom} = 10^{-10} \, \text{m}\):
\[ \lambda = 3 \times 10^{-10} \, \text{m} = 3 \, \text{angstrom} \]

### Final Answer
The de Broglie wavelength of the particle is \(3 \, \text{angstrom}\).

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