Question

The amount of work done to increase the speed of an electron from \( v = c/3 \) to \( v = 2c/3 \) is \[ c = 3 \times 10^8 \, \text{m/s}, \text{ rest mass of electron is } 0.511 \, \text{MeV} \]

• A. 56.50 keV
• B. 143.58 keV
• C. 168.20 keV
• D. 511.00 keV

Hint:

For relativistic speeds, use the Lorentz factor \( \gamma \) to calculate kinetic energy. The work done is the change in kinetic energy.

Answer 1

The Correct Option is B

Step 1: Relativistic Kinetic Energy Formula.
The relativistic kinetic energy is given by: \[ K.E. = (\gamma - 1) m c^2 \] where \( \gamma = \frac{1}{\sqrt{1 - (v/c)^2}} \), \( m \) is the rest mass, and \( c \) is the speed of light.
Step 2: Calculating \( \gamma \) at \( v = c/3 \) and \( v = 2c/3 \).
For \( v = c/3 \), \[ \gamma_1 = \frac{1}{\sqrt{1 - (1/3)^2}} = \frac{1}{\sqrt{8/9}} = \frac{3}{\sqrt{8}} \] For \( v = 2c/3 \), \[ \gamma_2 = \frac{1}{\sqrt{1 - (2/3)^2}} = \frac{1}{\sqrt{5/9}} = \frac{3}{\sqrt{5}} \]
Step 3: Work done.
The work done is the difference in kinetic energy: \[ W = (\gamma_2 - \gamma_1) m c^2 = \left( \frac{3}{\sqrt{5}} - \frac{3}{\sqrt{8}} \right) \times 0.511 \, \text{MeV} \] After evaluating the expression, we get \( W = 143.58 \, \text{keV} \).
Step 4: Conclusion.
Thus, the work done to increase the speed of the electron is 143.58 keV, so the correct answer is (B).