The velocity of electron of H-atom in its ground state is 1.4×`10^(5)`m/s. The de-Broglie wavelength of this electron would be:

Options

A. 3nm
B. 2nm
C. 5nm
D. none

Explanation

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

### Step 1: Understand the formula for de Broglie wavelength
The de Broglie wavelength (\(\lambda\)) 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.

### Step 2: Identify the known values
From the question, we have:
- Velocity of the electron, \(v = 1.4 \times 10^5 \, \text{m/s}\)
- Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- Mass of the electron, \(m = 9.1 \times 10^{-31} \, \text{kg}\)

### Step 3: Substitute the values into the formula
Now, we can substitute the known values into the de Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (1.4 \times 10^5)} \]

### Step 4: Calculate the denominator
First, calculate the denominator:
\[ (9.1 \times 10^{-31}) \times (1.4 \times 10^5) = 1.274 \times 10^{-25} \]

### Step 5: Calculate the de Broglie wavelength
Now, substitute the denominator back into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.274 \times 10^{-25}} \approx 5.19 \times 10^{-9} \, \text{m} \]

### Step 6: Convert to nanometers
To express the wavelength in nanometers, we convert meters to nanometers (1 m = \(10^9\) nm):
\[ \lambda \approx 5.19 \, \text{nm} \]

### Conclusion
The de Broglie wavelength of the electron in the hydrogen atom in its ground state is approximately \(5 \, \text{nm}\).