If the velocity of the electron in Bohr's first orbit is `2.19xx10^(6) m s^(-1)`, calculate the de Broglie wavelength associated with it.

Options

A. `2.19xx10^(-6)`m
B. `4.38xx10^(-6)`m
C. `3.32xx10^(-10)`m
D. `3.32xx10^(10)`m

Explanation

To calculate the de Broglie wavelength associated with the electron in Bohr's first orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the de Broglie wavelength formula**:
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. **Identify the values needed**:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Velocity of the electron \( v = 2.19 \times 10^{6} \, \text{m/s} \)

3. **Substitute the values into the formula**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \times (2.19 \times 10^{6})} \]

4. **Calculate the denominator**:
- First, calculate \( m \times v \):
\[ m \times v = 9.1 \times 10^{-31} \, \text{kg} \times 2.19 \times 10^{6} \, \text{m/s} = 1.993 \times 10^{-24} \, \text{kg m/s} \]

5. **Now calculate λ**:
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.993 \times 10^{-24}} \approx 3.32 \times 10^{-10} \, \text{m} \]

6. **Final Result**:
The de Broglie wavelength associated with the electron is:
\[ \lambda \approx 3.32 \times 10^{-10} \, \text{m} \]