The velocity of electron of H-atom in its ground state is 4.4×`10^(5)`m/s. The de-Broglie wavelength of this electron would be:

Options

A. 1.6nm
B. `6.1`nm
C. 11nm
D. None of these

Explanation

To find the de-Broglie wavelength of the electron in the ground state of a hydrogen atom, 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 (\(4.4 \times 10^{5} \, \text{m/s}\)).

### Step-by-Step Solution:

1. **Identify the given values**:
- Velocity of electron, \(v = 4.4 \times 10^{5} \, \text{m/s}\)
- Mass of electron, \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{Js}\)

2. **Substitute the values into the de-Broglie wavelength formula**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (4.4 \times 10^{5})} \]

3. **Calculate the denominator**:
- First, calculate \(m \times v\):
\[ m \times v = 9.1 \times 10^{-31} \, \text{kg} \times 4.4 \times 10^{5} \, \text{m/s} = 4.004 \times 10^{-25} \, \text{kg m/s} \]

4. **Now substitute back into the equation for \(\lambda\)**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{4.004 \times 10^{-25}} \]

5. **Perform the division**:
\[ \lambda \approx 1.65 \times 10^{-9} \, \text{m} \]

6. **Convert to nanometers**:
\[ \lambda \approx 1.65 \, \text{nm} \]

### Final Answer:
The de-Broglie wavelength of the electron in its ground state is approximately \(1.65 \, \text{nm}\).