Calculate de - Broglie wavelength of an electron having kinetic energy `2.8xx10^(-23)J`

Options

A. `9.28 xx10^(-4)m`
B. `9.28 xx10^(-7)m`
C. `9.28 xx10^(-8)m`
D. `9.28 xx10^(-10)m`

Answer

To calculate the de Broglie wavelength of an electron with a given kinetic energy, we can use the de Broglie wavelength formula:

\[ \lambda = \frac{h}{\sqrt{2mE}} \]

where:
- \(\lambda\) is the de Broglie wavelength,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(m\) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)),
- \(E\) is the kinetic energy of the electron.

### Step-by-Step Solution:

1. **Identify the values:**
- Kinetic energy \(E = 2.8 \times 10^{-23} \, \text{J}\)
- Planck's constant \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- Mass of the electron \(m = 9.1 \times 10^{-31} \, \text{kg}\)

2. **Substitute the values into the formula:**
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 2.8 \times 10^{-23}}} \]

3. **Calculate the denominator:**
- First, calculate \(2mE\):
\[ 2mE = 2 \times 9.1 \times 10^{-31} \times 2.8 \times 10^{-23} \]
\[ = 5.096 \times 10^{-53} \, \text{J kg} \]

4. **Calculate the square root:**
\[ \sqrt{2mE} = \sqrt{5.096 \times 10^{-53}} \approx 7.14 \times 10^{-27} \, \text{kg}^{1/2} \text{J}^{1/2} \]

5. **Calculate the wavelength \(\lambda\):**
\[ \lambda = \frac{6.626 \times 10^{-34}}{7.14 \times 10^{-27}} \approx 9.28 \times 10^{-8} \, \text{m} \]

### Final Answer:
The de Broglie wavelength of the electron is approximately \(9.28 \times 10^{-8} \, \text{m}\).