The velocity of electron of H-atom in its ground state is 2.4×`10^(5)`m/s. The de-Broglie wavelength of this electron would be:

Options

A. 3nm
B. 4nm
C. 6nm
D. 5nm

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 (\(2.4 \times 10^{5} \, \text{m/s}\)).

### Step-by-Step Solution:

1. **Identify the values needed for the calculation:**
- 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, \(v = 2.4 \times 10^{5} \, \text{m/s}\)

2. **Substitute the values into the de-Broglie wavelength formula:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (2.4 \times 10^{5})} \]

3. **Calculate the denominator:**
\[ mv = (9.1 \times 10^{-31}) \times (2.4 \times 10^{5}) = 2.184 \times 10^{-25} \, \text{kg m/s} \]

4. **Now, substitute the calculated value of \(mv\) back into the equation:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.184 \times 10^{-25}} \]

5. **Perform the division:**
\[ \lambda \approx 3.03 \times 10^{-9} \, \text{m} \]

6. **Convert meters to nanometers (1 m = \(10^{9}\) nm):**
\[ \lambda \approx 3.03 \, \text{nm} \]

### Final Answer:
The de-Broglie wavelength of the electron in the hydrogen atom in its ground state is approximately **3.03 nm**.