Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potentiation difference of 100 million volts
`[ 1eV = 1.6 xx 10^(-19)J, m_(e) = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) J s, c = 3.0 xx 10^(8) m s^(-1)]`

Explanation

To calculate the wavelength of an electron that has been accelerated through a potential difference of 100 million volts, we can use the de Broglie wavelength formula. Here are the steps to find the solution:

### Step 1: Understand the relationship between energy and potential difference
When an electron is accelerated through a potential difference (V), it gains kinetic energy equal to the electric potential energy. The energy (E) gained by the electron can be calculated using the formula:
\[ E = eV \]
where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{J} \))
- \( V \) is the potential difference in volts (100 million volts = \( 100 \times 10^6 \, \text{V} \))

### Step 2: Calculate the energy gained by the electron
Substituting the values:
\[ E = (1.6 \times 10^{-19} \, \text{J})(100 \times 10^6 \, \text{V}) \]
\[ E = 1.6 \times 10^{-19} \times 10^8 \]
\[ E = 1.6 \times 10^{-11} \, \text{J} \]

### Step 3: Relate energy to momentum
The kinetic energy of the electron can also be expressed in terms of its mass (m) and velocity (v):
\[ E = \frac{1}{2} mv^2 \]
From this, we can express the momentum (p) of the electron as:
\[ p = mv \]
Using the relationship between energy and momentum, we can derive:
\[ E = \frac{p^2}{2m} \]
This implies:
\[ p = \sqrt{2mE} \]

### Step 4: Substitute the values to find momentum
Now, we can substitute the values for mass (\( m = 9.1 \times 10^{-31} \, \text{kg} \)) and energy (\( E = 1.6 \times 10^{-11} \, \text{J} \)):
\[ p = \sqrt{2 \times (9.1 \times 10^{-31} \, \text{kg}) \times (1.6 \times 10^{-11} \, \text{J})} \]
Calculating this gives:
\[ p = \sqrt{2.912 \times 10^{-41}} \]
\[ p \approx 5.39 \times 10^{-21} \, \text{kg m/s} \]

### Step 5: Calculate the de Broglie wavelength
Now we can use the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)):
\[ \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{5.39 \times 10^{-21} \, \text{kg m/s}} \]
Calculating this gives:
\[ \lambda \approx 1.23 \times 10^{-12} \, \text{m} \]

### Step 6: Convert to nanometers
To convert meters to nanometers:
\[ \lambda \approx 1.23 \times 10^{-12} \, \text{m} = 0.123 \, \text{nm} \]

### Final Answer
The wavelength of the electron is approximately \( 0.123 \, \text{nm} \).

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