If the KE of electron is `2.5 xx 10^(-24)` J , then calculate its de-Broglie wavelength.

Options

A. 623.7 nm
B. 311.1 nm
C. 181.45 nm
D. 126.7 nm

Answer

To calculate the de-Broglie wavelength of an electron given its 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 \(m\) is the mass of the electron and \(v\) is its velocity.

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

### Step 3: Substitute the values
Given:
- \(KE = 2.5 \times 10^{-24} \, \text{J}\)
- Mass of the electron \(m = 9.1 \times 10^{-31} \, \text{kg}\)

Substituting these values into the equation for velocity:
\[ v = \sqrt{\frac{2 \cdot (2.5 \times 10^{-24})}{9.1 \times 10^{-31}}} \]

### Step 4: Calculate the velocity
Calculating the value inside the square root:
\[ v = \sqrt{\frac{5.0 \times 10^{-24}}{9.1 \times 10^{-31}}} \]
Calculating further:
\[ v = \sqrt{5.49 \times 10^{6}} \approx 2.34 \times 10^{3} \, \text{m/s} \]

### Step 5: Use the de-Broglie wavelength formula
The de-Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \(h\) is Planck's constant.

### Step 6: Substitute the values into the de-Broglie wavelength formula
Using:
- \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- \(v = 2.34 \times 10^{3} \, \text{m/s}\)

Substituting these values:
\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31})(2.34 \times 10^{3})} \]

### Step 7: Calculate the de-Broglie wavelength
Calculating the denominator:
\[ mv = (9.1 \times 10^{-31})(2.34 \times 10^{3}) \approx 2.13 \times 10^{-27} \, \text{kg m/s} \]
Now substituting back:
\[ \lambda = \frac{6.626 \times 10^{-34}}{2.13 \times 10^{-27}} \approx 3.11 \times 10^{-7} \, \text{m} \]

### Step 8: Convert to nanometers
To convert meters to nanometers:
\[ \lambda \approx 3.11 \times 10^{-7} \, \text{m} = 311.1 \, \text{nm} \]

### Final Answer
The de-Broglie wavelength of the electron is approximately \(311.1 \, \text{nm}\).

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