What will be de Broglie's wavelength of an electron moving with a velocity of `1.2 xx 10^(5) ms^(-1)` ?

Options

A. `6.068 xx 10^(-9)`
B. `3.133 xx 10^(-37)`
C. `6.626 xx 10^(-9)`
D. `6.018 xx 10^(-7)`

Explanation

To find the de Broglie wavelength of an electron moving with a velocity of \(1.2 \times 10^{5} \, \text{ms}^{-1}\), we will use the de Broglie wavelength formula:

\[ \lambda = \frac{h}{p} \]

where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is the Planck constant,
- \(p\) is the momentum of the electron.

### Step 1: Identify the values needed
1. **Planck constant (\(h\))**: \(6.626 \times 10^{-34} \, \text{Js}\)
2. **Mass of the electron (\(m_e\))**: \(9.109 \times 10^{-31} \, \text{kg}\)
3. **Velocity of the electron (\(v\))**: \(1.2 \times 10^{5} \, \text{ms}^{-1}\)

### Step 2: Calculate the momentum (\(p\))
Momentum (\(p\)) is calculated using the formula:

\[ p = m \cdot v \]

Substituting the values:

\[ p = (9.109 \times 10^{-31} \, \text{kg}) \cdot (1.2 \times 10^{5} \, \text{ms}^{-1}) \]

Calculating this gives:

\[ p = 1.09308 \times 10^{-25} \, \text{kg m/s} \]

### Step 3: Substitute the values into the de Broglie equation
Now, substitute \(h\) and \(p\) into the de Broglie equation:

\[ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \, \text{Js}}{1.09308 \times 10^{-25} \, \text{kg m/s}} \]

### Step 4: Calculate the wavelength (\(\lambda\))
Performing the division:

\[ \lambda = 6.065 \times 10^{-9} \, \text{m} \]

### Final Answer
Thus, the de Broglie wavelength of the electron is approximately:

\[ \lambda \approx 6.065 \times 10^{-9} \, \text{m} \text{ or } 6.065 \, \text{nm} \]

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