The velocity of electron of H-atom in its ground state is 1.4×`10^(6)`m/s. The de-Broglie wavelength of this electron would be:

Options

A. 0.51nm
B. 0.15nm
C. 0.25nm
D. 0.17nm

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 the 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 (\(1.4 \times 10^{6} \, \text{m/s}\)).

### Step-by-step Solution:

1. **Identify the given values:**
- Velocity of the electron, \(v = 1.4 \times 10^{6} \, \text{m/s}\)
- Mass of the 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:**
\[ \lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (1.4 \times 10^{6})} \]

3. **Calculate the denominator:**
\[ mv = (9.1 \times 10^{-31} \, \text{kg}) \times (1.4 \times 10^{6} \, \text{m/s}) = 1.274 \times 10^{-24} \, \text{kg m/s} \]

4. **Calculate the wavelength:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.274 \times 10^{-24}} \approx 5.20 \times 10^{-10} \, \text{m} \]

5. **Convert meters to nanometers:**
\[ \lambda \approx 0.52 \, \text{nm} \quad \text{(since } 1 \, \text{nm} = 10^{-9} \, \text{m)} \]

### Final Answer:
The de-Broglie wavelength of the electron in the hydrogen atom in its ground state is approximately \(0.52 \, \text{nm}\).