The wave length of a electron with mass `9.1 xx 10^(-31)kg` and kinetic energy `3.0 xx 10^(-25)J` is

Options

A. `89.67nm`
B. `8.96nm`
C. `456.7nm`
D. `896.7nm`

Explanation

To find the wavelength of an electron given its mass and kinetic energy, we can use the de Broglie wavelength formula and the kinetic energy equation. Here’s a step-by-step solution:

### Step 1: Calculate the velocity of the electron
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.

We can rearrange this formula to solve for \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m}} \]

### Step 2: Substitute the values into the equation
Given:
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Kinetic energy, \( KE = 3.0 \times 10^{-25} \, \text{J} \)

Substituting these values into the equation for \( v \):
\[ v = \sqrt{\frac{2 \times (3.0 \times 10^{-25})}{9.1 \times 10^{-31}}} \]

### Step 3: Calculate the velocity
Calculating the numerator:
\[ 2 \times 3.0 \times 10^{-25} = 6.0 \times 10^{-25} \]
Now, divide by the mass:
\[ \frac{6.0 \times 10^{-25}}{9.1 \times 10^{-31}} \approx 6.59341 \times 10^{5} \]
Now take the square root:
\[ v \approx \sqrt{6.59341 \times 10^{5}} \approx 811.58 \, \text{m/s} \]

### Step 4: Use the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
Where:
- \( h \) is Planck's constant, \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( p \) is the momentum, which can be calculated as \( p = mv \)

### Step 5: Calculate the momentum
Using the mass and the velocity calculated:
\[ p = mv = (9.1 \times 10^{-31}) \times (811.58) \approx 7.394 \times 10^{-28} \, \text{kg m/s} \]

### Step 6: Calculate the wavelength
Now substitute \( p \) back into the de Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{7.394 \times 10^{-28}} \approx 8.96 \times 10^{-7} \, \text{m} \]

### Step 7: Convert to nanometers
Since \( 1 \, \text{nm} = 10^{-9} \, \text{m} \):
\[ \lambda \approx 8.96 \times 10^{-7} \, \text{m} = 896 \, \text{nm} \]

### Final Answer:
The wavelength of the electron is approximately **896 nm**.

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