Einstein in `1905` proppunded the special theory of relativity and in `1915` proposed the general theory of relativity. The special theory deals with inertial frames of reference. The general theory of relativity deals with problems in which one frame of reference. He assumed that fixed frame is accelerated w.r.t. another frame of reference of reference cannot be located. Postulated of special theory of realtivity
● The laws of physics have the same form in all inertial systems.
● The velocity light in empty space is a unicersal constant the same for all observers.
● Einstein proved the following facts based on his theory of special relativity. Let v be the velocity of the speceship w.r.t a given frame of reference. The obserations are made by an observer in that reference frame.
● All clocks on the spaceship wil go slow by a factor `sqrt(1-v^(2)//c^(2))`
● All objects on the spaceship will have contracted in length by a factor `sqrt(1-v^(2)//c^(2))`
● The mass of the spaceship increases by a factor `sqrt(1-v^(2)//c^(2))`
● Mass and energy are interconvertable `E = mc^(2)`
The speed of a meterial object can never exceed the velocity of light.
● If two objects A and B are moving with velocity u and v w.r.t each other along the `x`-axis, the relative velocity of A w.r.t. `B = (u-v)/(1-uv//v^(2))`
The momentum of an electron moving with a speed `0.6 c` (Rest mass of electron is `9.1 xx 10^(-31 kg`)

Options

A. `1.6 xx 10^(-22) kgms^(-1)`
B. `2 xx 10^(-22) kgms^(-1)`
C. `5.46 xx 10^(-31) kgms^(-1)`
D. `5.46 xx 10^(-22) kgms^(-1)`

Explanation

To solve the problem of finding the momentum of an electron moving at a speed of \(0.6c\) (where \(c\) is the speed of light), we will follow these steps:

### Step 1: Understand the Concept of Relativistic Momentum
In special relativity, the momentum \(P\) of an object moving at a significant fraction of the speed of light is given by the formula:
\[ P = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \]
where:
- \(m_0\) is the rest mass of the object,
- \(v\) is the velocity of the object,
- \(c\) is the speed of light.

### Step 2: Identify Given Values
From the problem, we have:
- The rest mass of the electron, \(m_0 = 9.1 \times 10^{-31} \, \text{kg}\),
- The speed of the electron, \(v = 0.6c\),
- The speed of light, \(c = 3 \times 10^8 \, \text{m/s}\).

### Step 3: Substitute Values into the Momentum Formula
Substituting the known values into the momentum formula:
\[ P = \frac{9.1 \times 10^{-31} \, \text{kg} \times (0.6 \times 3 \times 10^8 \, \text{m/s})}{\sqrt{1 - \left(0.6 \times 3 \times 10^8 \, \text{m/s}\right)^2 / (3 \times 10^8 \, \text{m/s})^2}} \]

### Step 4: Calculate the Velocity
Calculating \(0.6c\):
\[ 0.6c = 0.6 \times 3 \times 10^8 = 1.8 \times 10^8 \, \text{m/s} \]

### Step 5: Calculate the Denominator
Now, calculate the denominator:
\[ \sqrt{1 - \frac{(1.8 \times 10^8)^2}{(3 \times 10^8)^2}} = \sqrt{1 - \frac{3.24 \times 10^{16}}{9 \times 10^{16}}} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 \]

### Step 6: Substitute Back into the Momentum Equation
Now substitute back into the momentum equation:
\[ P = \frac{9.1 \times 10^{-31} \times 1.8 \times 10^8}{0.8} \]

### Step 7: Calculate the Momentum
Calculating the numerator:
\[ 9.1 \times 10^{-31} \times 1.8 \times 10^8 = 1.638 \times 10^{-22} \, \text{kg m/s} \]
Now divide by \(0.8\):
\[ P = \frac{1.638 \times 10^{-22}}{0.8} = 2.0475 \times 10^{-22} \, \text{kg m/s} \]

### Step 8: Final Result
Thus, the momentum of the electron is approximately:
\[ P \approx 2 \times 10^{-22} \, \text{kg m/s} \]

### Conclusion
The correct answer is option number B.

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