Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential differecne of 1kV.

Answer

To calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 kV, we can follow these steps:

### Step 1: Calculate the Kinetic Energy of the Electron
The kinetic energy (KE) of an electron accelerated through a potential difference (V) is given by the formula:

\[ KE = e \cdot V \]

Where:
- \( e \) is the charge of the electron, approximately \( 1.6 \times 10^{-19} \) coulombs.
- \( V \) is the potential difference in volts (1 kV = 1000 V).

Substituting the values:

\[ KE = (1.6 \times 10^{-19} \, \text{C}) \cdot (1000 \, \text{V}) = 1.6 \times 10^{-16} \, \text{J} \]

### Step 2: Relate Kinetic Energy to Velocity
The kinetic energy can also be expressed in terms of the mass (m) and velocity (v) of the electron:

\[ KE = \frac{1}{2} m v^2 \]

Setting the two expressions for kinetic energy equal gives:

\[ \frac{1}{2} m v^2 = e \cdot V \]

Rearranging for velocity \( v \):

\[ v = \sqrt{\frac{2 e V}{m}} \]

### Step 3: Substitute Values to Calculate Velocity
The mass of the electron \( m \) is approximately \( 9.1 \times 10^{-31} \) kg. Substituting the known values into the equation:

\[ v = \sqrt{\frac{2 \cdot (1.6 \times 10^{-19} \, \text{C}) \cdot (1000 \, \text{V})}{9.1 \times 10^{-31} \, \text{kg}}} \]

Calculating the value inside the square root:

\[ v = \sqrt{\frac{3.2 \times 10^{-16}}{9.1 \times 10^{-31}}} \]

Calculating this gives:

\[ v \approx \sqrt{3.51 \times 10^{14}} \approx 1.875 \times 10^7 \, \text{m/s} \]

### Step 4: Calculate the de Broglie Wavelength
The de Broglie wavelength \( \lambda \) is given by the formula:

\[ \lambda = \frac{h}{m v} \]

Where:
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{kg m}^2/\text{s} \).

Substituting the values:

\[ \lambda = \frac{6.626 \times 10^{-34}}{(9.1 \times 10^{-31}) \cdot (1.875 \times 10^7)} \]

Calculating the denominator:

\[ \lambda = \frac{6.626 \times 10^{-34}}{1.707 \times 10^{-23}} \approx 3.88 \times 10^{-11} \, \text{m} \]

### Final Answer
The de Broglie wavelength of the electron is approximately:

\[ \lambda \approx 3.88 \times 10^{-11} \, \text{m} \text{ or } 0.388 \, \text{nm} \]

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