The wavelength of de - Broglie wave is `2 mu m` , then its momentum is `( h = 6.63 xx 10^(-34 ) J-s)`

Options

A. `3.315 xx10^(-28) kg ms ^(-1)`
B. `1.66 xx10^(-28) kg ms ^(-1)`
C. `4.97 xx10^(-28) kg ms ^(-1)`
D. `9.9 xx10^(-28) kg ms ^(-1)`

Answer

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

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

Where:
- \(\lambda\) is the wavelength,
- \(h\) is Planck's constant,
- \(p\) is the momentum.

### Step 1: Convert the Wavelength to Meters
The given wavelength is \(2 \mu m\). We need to convert this to meters:

\[ \lambda = 2 \mu m = 2 \times 10^{-6} \text{ m} \]

### Step 2: Rearrange the Formula to Solve for Momentum
We want to find the momentum \(p\). Rearranging the formula gives:

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

### Step 3: Substitute the Values
We know:
- \(h = 6.63 \times 10^{-34} \text{ J-s}\)
- \(\lambda = 2 \times 10^{-6} \text{ m}\)

Now substituting these values into the equation:

\[ p = \frac{6.63 \times 10^{-34}}{2 \times 10^{-6}} \]

### Step 4: Perform the Calculation
Calculating the momentum:

\[ p = \frac{6.63}{2} \times 10^{-34 + 6} = 3.315 \times 10^{-28} \text{ kg m/s} \]

### Final Answer
The momentum of the particle is:

\[ p = 3.315 \times 10^{-28} \text{ kg m/s} \]