The velocity of electron of H-atom in its ground state is 8.4×`10^(5)`m/s. The de-Broglie wavelength of this electron would be:

Options

A. 0.86nm
B. 6.8nm
C. 0.48nm
D. 8.4nm

Explanation

To find the de Broglie wavelength of the electron in a hydrogen atom in its ground state, we can use the de Broglie wavelength formula:

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

where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)),
- \(v\) is the velocity of the electron (\(8.4 \times 10^{5} \, \text{m/s}\)).

### Step-by-step Solution:

1. **Identify the constants and 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, \(v = 8.4 \times 10^{5} \, \text{m/s}\)

2. **Substitute the values into the de Broglie wavelength formula**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (8.4 \times 10^{5})} \]

3. **Calculate the denominator**:
- First, calculate \(m \times v\):
\[ m \times v = 9.1 \times 10^{-31} \times 8.4 \times 10^{5} \approx 7.644 \times 10^{-25} \, \text{kg m/s} \]

4. **Now, calculate the wavelength**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{7.644 \times 10^{-25}} \approx 8.68 \times 10^{-10} \, \text{m} \]

5. **Convert the wavelength to nanometers**:
- Since \(1 \, \text{nm} = 10^{-9} \, \text{m}\):
\[ \lambda \approx 0.868 \, \text{nm} \approx 0.86 \, \text{nm} \]

### Final Answer:
The de Broglie wavelength of the electron in the hydrogen atom in its ground state is approximately **0.86 nanometers**.