The velocity of electron of H-atom in its ground state is 4.2×`10^(6)`m/s. The de-Broglie wavelength of this electron would be:

Options

A. 23.3 nm
B. .17 nm
C. 2.3 nm
D. 14.7nm

Explanation

To find the de Broglie wavelength of the electron in a hydrogen atom in its ground state, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the formula for de Broglie wavelength**:
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron.

2. **Gather 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 \( v = 4.2 \times 10^{6} \, \text{m/s} \)

3. **Substitute the values into the formula**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (4.2 \times 10^{6})} \]

4. **Calculate the denominator**:
- First, calculate \( m \times v \):
\[ m \times v = 9.1 \times 10^{-31} \times 4.2 \times 10^{6} = 3.822 \times 10^{-24} \, \text{kg m/s} \]

5. **Calculate the de Broglie wavelength**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{3.822 \times 10^{-24}} \approx 1.73 \times 10^{-10} \, \text{m} \]

6. **Convert the wavelength to nanometers**:
\[ 1.73 \times 10^{-10} \, \text{m} = 0.173 \, \text{nm} \]

### Final Answer:
The de Broglie wavelength of the electron in the hydrogen atom in its ground state is approximately \( 0.173 \, \text{nm} \).

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