Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potential difference of 1 kv. `[ " Given 1 eV"=1.6 xx 10^(-19)J]`

Answer

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

### Step 1: Determine the kinetic energy of the electron
The kinetic energy (KE) gained by the electron when it is accelerated through a potential difference (V) is given by the equation:
\[ KE = e \cdot V \]
Where:
- \( e \) is the charge of the electron, which is approximately \( 1.6 \times 10^{-19} \, \text{C} \).
- \( V \) 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:
\[ KE = \frac{1}{2} mv^2 \]

We can rearrange this equation to solve for the velocity (v):
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

The mass of the electron (m) is approximately \( 9.11 \times 10^{-31} \, \text{kg} \).

Substituting the values:
\[ v = \sqrt{\frac{2 \cdot (1.6 \times 10^{-16} \, \text{J})}{9.11 \times 10^{-31} \, \text{kg}}} \]

Calculating this gives:
\[ v \approx \sqrt{\frac{3.2 \times 10^{-16}}{9.11 \times 10^{-31}}} \]
\[ v \approx \sqrt{3.51 \times 10^{14}} \]
\[ v \approx 5.93 \times 10^{7} \, \text{m/s} \]

### Step 3: Calculate the wavelength using the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) of a particle is given by:
\[ \lambda = \frac{h}{mv} \]
Where:
- \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \, \text{Js} \).

Substituting the values:
\[ \lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{(9.11 \times 10^{-31} \, \text{kg}) \cdot (5.93 \times 10^{7} \, \text{m/s})} \]

Calculating this gives:
\[ \lambda = \frac{6.63 \times 10^{-34}}{5.39 \times 10^{-23}} \]
\[ \lambda \approx 1.23 \times 10^{-11} \, \text{m} \]
or
\[ \lambda \approx 0.0123 \, \text{nm} \]

### Final Answer
The wavelength of the electron accelerated through a potential difference of 1 kV is approximately \( 0.0123 \, \text{nm} \).

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