An electron has kinetic energy `2.8 xx 10^-23 J` de-Broglie wavelength will be nearly.
`(m_e = 9.1 xx 10^-31 kg)`.

Options

A. `9.28 xx 10^-4 m`
B. `9.28 xx 10^-7 m`
C. `9.28 xx 10^-8 m`
D. `9.28 xx 10^-10 m`

Explanation

To find the de Broglie wavelength of an electron given its kinetic energy, we can follow these steps:

### Step 1: Write down the given values
- Kinetic Energy (KE) of the electron, \( KE = 2.8 \times 10^{-23} \, J \)
- Mass of the electron, \( m_e = 9.1 \times 10^{-31} \, kg \)
- Planck's constant, \( h = 6.626 \times 10^{-34} \, J \cdot s \)

### Step 2: Use the kinetic energy formula to find the velocity of the electron
The kinetic energy of an electron can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Rearranging this formula to find the velocity \( v \):
\[ v^2 = \frac{2 \times KE}{m} \]
Substituting the known values:
\[ v^2 = \frac{2 \times 2.8 \times 10^{-23}}{9.1 \times 10^{-31}} \]

### Step 3: Calculate \( v^2 \)
Calculating the right side:
\[ v^2 = \frac{5.6 \times 10^{-23}}{9.1 \times 10^{-31}} \approx 6.15 \times 10^{7} \]
Now, taking the square root to find \( v \):
\[ v \approx \sqrt{6.15 \times 10^{7}} \approx 7.81 \times 10^{3} \, m/s \]

### Step 4: Calculate the de Broglie wavelength
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
Substituting the known values:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31})(7.81 \times 10^{3})} \]

### Step 5: Calculate \( \lambda \)
Calculating the denominator:
\[ mv = (9.1 \times 10^{-31})(7.81 \times 10^{3}) \approx 7.10 \times 10^{-27} \]
Now substituting back into the wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{7.10 \times 10^{-27}} \approx 9.28 \times 10^{-8} \, m \]

### Final Answer
The de Broglie wavelength of the electron is approximately:
\[ \lambda \approx 9.28 \times 10^{-8} \, m \]

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