The velocity of electron of H-atom in its ground state is 2.2×`10^(−6)`m/s. The de-Broglie wavelength of this electron would be:

Options

A. 0.33 nm
B. 23.3 nm
C. 45.6 nm
D. 100nm

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 (\(2.2 \times 10^{6} \, \text{m/s}\)).

### Step-by-Step Solution:

1. **Identify the known 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, \(v = 2.2 \times 10^{6} \, \text{m/s}\)

2. **Substitute the values into the de Broglie wavelength formula:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (2.2 \times 10^{6})} \]

3. **Calculate the denominator:**
- First, calculate \(m \times v\):
\[ m \times v = 9.1 \times 10^{-31} \times 2.2 \times 10^{6} = 2.002 \times 10^{-24} \, \text{kg m/s} \]

4. **Now calculate the de Broglie wavelength:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.002 \times 10^{-24}} \approx 3.31 \times 10^{-10} \, \text{m} \]

5. **Convert the wavelength to nanometers:**
- Since \(1 \, \text{nm} = 10^{-9} \, \text{m}\):
\[ \lambda \approx 0.331 \, \text{nm} \]

Thus, the de Broglie wavelength of the electron in the hydrogen atom in its ground state is approximately \(0.331 \, \text{nm}\).

### Final Answer:
\[ \lambda \approx 0.331 \, \text{nm} \]