The momentum of a paricle which has a d-Broglie wavelength of 0.1 nm is `( h= 6.6 xx10^(-34) J s)`

Options

A. `3.2 xx10^(-24) kg ms^(-1)`
B. ` 4.3 xx10^(-22) kgms^(-1)`
C. ` 5.3 xx 10^(-22) kgms^(-1)`
D. ` 6.62 xx 10^(-24) kg ms^(-1)`

Answer

To find the momentum of a particle with a de Broglie wavelength of 0.1 nm, we can follow these steps:

### Step 1: Convert the wavelength from nanometers to meters
The given wavelength (λ) is 0.1 nm. We know that:
1 nm = \(10^{-9}\) meters.

Thus,
\[ 0.1 \text{ nm} = 0.1 \times 10^{-9} \text{ m} = 10^{-10} \text{ m} \]

### Step 2: Use the de Broglie wavelength formula
The de Broglie wavelength formula relates the wavelength (λ) to the momentum (p) of a particle:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant.

### Step 3: Rearrange the formula to solve for momentum
Rearranging the formula to find momentum (p):
\[ p = \frac{h}{\lambda} \]

### Step 4: Substitute the values into the equation
We know:
- \(h = 6.626 \times 10^{-34} \text{ J s}\)
- \(\lambda = 10^{-10} \text{ m}\)

Substituting these values into the momentum equation:
\[ p = \frac{6.626 \times 10^{-34} \text{ J s}}{10^{-10} \text{ m}} \]

### Step 5: Calculate the momentum
Now, perform the calculation:
\[ p = 6.626 \times 10^{-34} \div 10^{-10} = 6.626 \times 10^{-24} \text{ kg m/s} \]

### Step 6: Write the final answer
The momentum of the particle is:
\[ p = 6.626 \times 10^{-24} \text{ kg m/s} \]

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