The wavelength associated with an electron moving with velocity `10^(10) ms^(-1)` is

Options

A. `6.62 x10^(-10) m`
B. ` 7.27 xx 10^(-14) m`
C. ` 3.69 xx 10^(-12) m`
D. ` 4.92 xx 10^(-11) m`

Answer

To find the wavelength associated with an electron moving with a velocity of \(10^{10} \, \text{m/s}\), we will use the de Broglie wavelength formula:

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

where:
- \(\lambda\) = wavelength
- \(h\) = Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))
- \(m\) = mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\))
- \(v\) = velocity of the electron (\(10^{10} \, \text{m/s}\))

### Step-by-step Solution:

1. **Identify the values**:
- Planck's constant \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- Mass of the electron \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- Velocity of the electron \(v = 10^{10} \, \text{m/s}\)

2. **Substitute the values into the de Broglie equation**:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{(9.1 \times 10^{-31} \, \text{kg})(10^{10} \, \text{m/s})} \]

3. **Calculate the denominator**:
- First, calculate \(m \cdot v\):
\[ m \cdot v = 9.1 \times 10^{-31} \, \text{kg} \times 10^{10} \, \text{m/s} = 9.1 \times 10^{-21} \, \text{kg m/s} \]

4. **Now substitute back into the equation**:
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{Js}}{9.1 \times 10^{-21} \, \text{kg m/s}} \]

5. **Perform the division**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-21}} \approx 7.27 \times 10^{-14} \, \text{m} \]

6. **Final result**:
The wavelength associated with the electron is:
\[ \lambda \approx 7.27 \times 10^{-14} \, \text{m} \]