The momentum of a particle having a de-Broglie wavelength of `10^(17)` m is (Given, `h=6.625xx10^(-34)m`)

Options

A. `3.3125xx10^(-7) kg ms^(-1)`
B. `26.5xx10^(-7) kg m s^(-1)`
C. `6.625xx10^(-17) kg ms^(-1)`
D. `13.25xx10^(-17) kg ms^(-1)`

Explanation

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

### Step 1: Write the de-Broglie wavelength formula
The de-Broglie wavelength (\( \lambda \)) is related to the momentum (\( p \)) of a particle by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant.

### Step 2: Rearrange the formula to find momentum
We can rearrange the formula to solve for momentum:
\[ p = \frac{h}{\lambda} \]

### Step 3: Substitute the given values
We know:
- \( h = 6.625 \times 10^{-34} \, \text{Joule} \cdot \text{second} \)
- \( \lambda = 10^{-17} \, \text{m} \)

Now substituting these values into the formula:
\[ p = \frac{6.625 \times 10^{-34}}{10^{-17}} \]

### Step 4: Perform the calculation
Calculating the momentum:
\[ p = 6.625 \times 10^{-34} \times 10^{17} = 6.625 \times 10^{-17} \, \text{kg} \cdot \text{m/s} \]

### Step 5: State the final answer
Thus, the momentum of the particle is:
\[ p = 6.625 \times 10^{-17} \, \text{kg} \cdot \text{m/s} \]