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 with a given kinetic energy, we can follow these steps:

### Step 1: Write down the formula for kinetic energy
The kinetic energy (KE) of an electron is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( KE \) is the kinetic energy,
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron.

### Step 2: Rearrange the formula to solve for velocity
We can rearrange the formula to solve for \( v \):
\[ v^2 = \frac{2 \cdot KE}{m} \]
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

### Step 3: Substitute the values into the equation
Given:
- \( KE = 2.8 \times 10^{-23} \, J \)
- \( m = 9.1 \times 10^{-31} \, kg \)

Substituting the values:
\[ v = \sqrt{\frac{2 \cdot (2.8 \times 10^{-23})}{9.1 \times 10^{-31}}} \]

### Step 4: Calculate the velocity
Calculating inside the square root:
\[ v = \sqrt{\frac{5.6 \times 10^{-23}}{9.1 \times 10^{-31}}} \]
\[ v = \sqrt{6.15 \times 10^7} \]
\[ v \approx 7.81 \times 10^3 \, m/s \]

### Step 5: Use the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) is given by:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \, J \cdot s \).

### Step 6: Substitute the values into the de Broglie wavelength formula
Substituting the values:
\[ \lambda = \frac{6.63 \times 10^{-34}}{(9.1 \times 10^{-31}) \cdot (7.81 \times 10^3)} \]

### Step 7: Calculate the de Broglie wavelength
Calculating the denominator:
\[ \lambda = \frac{6.63 \times 10^{-34}}{7.1 \times 10^{-27}} \]
\[ \lambda \approx 9.28 \times 10^{-8} \, m \]

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

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