If the Planck’s constant `h = 6.6xx10^(–34) Js`, the de Broglie wavelength of a particle having momentum of `3.3 xx 10^(-24) kg ms^(-1)` will be

Options

A. `0.002 Å`
B. `0.5 Å`
C. `2 Å`
D. `500 Å`

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 given values
- Planck's constant, \(h = 6.6 \times 10^{-34} \, \text{Js}\)
- Momentum, \(p = 3.3 \times 10^{-24} \, \text{kg m/s}\)

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

\[ \lambda = \frac{6.6 \times 10^{-34} \, \text{Js}}{3.3 \times 10^{-24} \, \text{kg m/s}} \]

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

\[ \lambda = \frac{6.6}{3.3} \times \frac{10^{-34}}{10^{-24}} \, \text{m} \]

Calculating the numerical part:

\[ \frac{6.6}{3.3} = 2 \]

And for the powers of ten:

\[ \frac{10^{-34}}{10^{-24}} = 10^{-34 + 24} = 10^{-10} \]

So we combine these results:

\[ \lambda = 2 \times 10^{-10} \, \text{m} \]

### Step 4: Convert to Angstroms
To convert meters to Angstroms (1 Angstrom = \(10^{-10}\) m):

\[ \lambda = 2 \times 10^{-10} \, \text{m} = 2 \, \text{Å} \]

### Final Answer
The de Broglie wavelength of the particle is:

\[ \lambda = 2 \, \text{Å} \]