An electron gun inside a computer monitor sends an electron toward the screen at a speed of `1.20xx10^(8)m//s`. If the mass of the electron is `9.190xx10^(-31)`kg, what is the magnitude of its relativistic momentum?

Options

A. `9.88xx10^(-23)kg*m//s`
B. `1.19xx10^(-22)kg*m//s`
C. `1.09xx10^(-22)kg*m//s`
D. `1.41xx10^(-22)kg*m//s`

Answer

To calculate the relativistic momentum of an electron moving at a speed of \(1.20 \times 10^8 \, \text{m/s}\), we will follow these steps:

### Step 1: Identify the known values
- Speed of the electron, \(v = 1.20 \times 10^8 \, \text{m/s}\)
- Mass of the electron, \(m = 9.19 \times 10^{-31} \, \text{kg}\)
- Speed of light, \(c = 3.00 \times 10^8 \, \text{m/s}\)

### Step 2: Calculate the relativistic factor \(\gamma\)
The relativistic factor \(\gamma\) is given by the formula:
\[ \gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \]
Substituting the values:
\[ \frac{v}{c} = \frac{1.20 \times 10^8}{3.00 \times 10^8} = 0.4 \]
Now, calculate \(\left(\frac{v}{c}\right)^2\):
\[ \left(\frac{v}{c}\right)^2 = (0.4)^2 = 0.16 \]
Now, calculate \(\gamma\):
\[ \gamma = \frac{1}{\sqrt{1 - 0.16}} = \frac{1}{\sqrt{0.84}} \approx 1.087 \]

### Step 3: Calculate the classical momentum
Classical momentum \(p\) is given by:
\[ p = mv \]
Substituting the values:
\[ p = (9.19 \times 10^{-31} \, \text{kg})(1.20 \times 10^8 \, \text{m/s}) = 1.103 \times 10^{-22} \, \text{kg m/s} \]

### Step 4: Calculate the relativistic momentum
The relativistic momentum \(p_{rel}\) is given by:
\[ p_{rel} = \gamma mv \]
Substituting the values:
\[ p_{rel} = (1.087)(9.19 \times 10^{-31} \, \text{kg})(1.20 \times 10^8 \, \text{m/s}) \]
Calculating this:
\[ p_{rel} \approx 1.087 \times 1.103 \times 10^{-22} \approx 1.198 \times 10^{-22} \, \text{kg m/s} \]

### Final Result
The magnitude of the relativistic momentum of the electron is approximately:
\[ p_{rel} \approx 1.198 \times 10^{-22} \, \text{kg m/s} \]

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