Find the momentum of an electron having wavelength `2A^(0)(h=6.62xx10^(-34)Js)`

Options

A. `1xx10^(-24)kg ms^(-1)`
B. `3.3xx10^(-24)kg ms^(-1)`
C. `4xx10^(-23)kg ms^(-1)`
D. `6xx10^(-24)kg ms^(-1)`

Answer

To find the momentum of an electron with a given wavelength, we can use the formula that relates momentum (p) to wavelength (λ) and Planck's constant (h). The formula is:

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Wavelength (λ) = 2 Å (Angstrom)
- Planck's constant (h) = \(6.62 \times 10^{-34} \, \text{Js}\)

2. **Convert the wavelength from Angstroms to meters:**
- 1 Å = \(10^{-10}\) meters
- Therefore, \(2 \, \text{Å} = 2 \times 10^{-10} \, \text{m}\)

3. **Substitute the values into the momentum formula:**
- Using the formula \( p = \frac{h}{\lambda} \):
\[ p = \frac{6.62 \times 10^{-34} \, \text{Js}}{2 \times 10^{-10} \, \text{m}} \]

4. **Calculate the momentum:**
- Performing the division:
\[ p = \frac{6.62 \times 10^{-34}}{2 \times 10^{-10}} = 3.31 \times 10^{-24} \, \text{kg m/s} \]

5. **Final answer:**
- The momentum of the electron is approximately \(3.31 \times 10^{-24} \, \text{kg m/s}\).