The velocity of electron of H-atom in its ground state is 3.2×`10^(6)`m/s. The de-Broglie wavelength of this electron would be:

Options

A. .22nm
B. 2.2nm
C. 32nm
D. 22nm

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 (\( 3.2 \times 10^{6} \, \text{m/s} \)).

### Step 1: Identify the values
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( v = 3.2 \times 10^{6} \, \text{m/s} \)

### Step 2: Substitute the values into the formula
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (3.2 \times 10^{6})} \]

### Step 3: Calculate the denominator
\[ 9.1 \times 10^{-31} \times 3.2 \times 10^{6} = 2.912 \times 10^{-24} \]

### Step 4: Calculate the wavelength
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.912 \times 10^{-24}} \approx 2.27 \times 10^{-10} \, \text{m} \]

### Step 5: Convert to nanometers
Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \):
\[ \lambda \approx 0.227 \, \text{nm} \]

### Final Answer
The de-Broglie wavelength of the electron in the hydrogen atom in its ground state is approximately \( 0.227 \, \text{nm} \).