The de broglie wavelength of an electron moving with a speed of `6.6xx10^5 m//s` is of the order of `(h=6.6xx10^(-34) Js " and " m_e = 9xx10^(-31) kg)`

Options

A. `10^(-12) m`
B. `10^(-11) m`
C. `10^(-9) m`
D. `10^(-7) m`

Explanation

To find the de Broglie wavelength of an electron moving with a speed of \(6.6 \times 10^5 \, \text{m/s}\), we can use the de Broglie wavelength formula:

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

where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is the Planck constant (\(6.6 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the electron (\(9 \times 10^{-31} \, \text{kg}\)),
- \(v\) is the speed of the electron (\(6.6 \times 10^5 \, \text{m/s}\)).

### Step-by-Step Solution:

1. **Identify the given values:**
- Planck constant, \(h = 6.6 \times 10^{-34} \, \text{Js}\)
- Mass of electron, \(m = 9 \times 10^{-31} \, \text{kg}\)
- Speed of electron, \(v = 6.6 \times 10^5 \, \text{m/s}\)

2. **Substitute the values into the de Broglie wavelength formula:**
\[ \lambda = \frac{6.6 \times 10^{-34}}{(9 \times 10^{-31}) \times (6.6 \times 10^5)} \]

3. **Calculate the denominator:**
\[ mv = (9 \times 10^{-31}) \times (6.6 \times 10^5) = 5.94 \times 10^{-25} \, \text{kg m/s} \]

4. **Now substitute this value back into the equation for \(\lambda\):**
\[ \lambda = \frac{6.6 \times 10^{-34}}{5.94 \times 10^{-25}} \]

5. **Perform the division:**
\[ \lambda \approx 1.11 \times 10^{-9} \, \text{m} \]

6. **Express the wavelength in scientific notation:**
\[ \lambda \approx 1.1 \times 10^{-9} \, \text{m} \]

7. **Determine the order of the de Broglie wavelength:**
The order of the wavelength is \(10^{-9} \, \text{m}\).

### Conclusion:
The de Broglie wavelength of the electron moving at the given speed is of the order of \(10^{-9} \, \text{m}\).