Question

Given that the rest mass of electron is 0.511MeV/c ² , the speed (in units of c) of an electron with kinetic energy 5.11MeV is closest to:

• A. 0.996
• B. 0.993
• C. 0.990
• D. 0.998

Answer 1

The Correct Option is A

We can use the relativistic energy-momentum relation to find the speed of the electron. The total energy \( E \) of a particle is given by:

\[ E = \gamma m c^2 \]

where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \) is the Lorentz factor, \( m \) is the rest mass, and \( v \) is the speed of the particle.

Step 1: Relate Total Energy to Kinetic Energy

The total energy is the sum of the rest mass energy and the kinetic energy:

\[ E = m c^2 + \text{K.E.} \]

Substitute the given values:

• Rest mass energy: \( m c^2 = 0.511 \, \text{MeV} \)
• Kinetic energy: \( \text{K.E.} = 5.11 \, \text{MeV} \)

Thus, the total energy is:

\[ E = 0.511 + 5.11 = 5.621 \, \text{MeV} \]

Step 2: Solve for \( \gamma \)

Using the formula \( E = \gamma m c^2 \), solve for \( \gamma \):

\[ \gamma = \frac{E}{m c^2} = \frac{5.621}{0.511} \approx 11 \]

Step 3: Solve for Speed \( v \)

From the Lorentz factor \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \), rearrange to find \( v \):

\[ v = c \sqrt{1 - \frac{1}{\gamma^2}} \]

Substitute \( \gamma = 11 \):

\[ v = c \sqrt{1 - \frac{1}{11^2}} = c \sqrt{1 - 0.00826} \approx c \sqrt{0.99174} \]

\[ v \approx 0.996c \]

Conclusion:

Thus, the speed of the electron is closest to \( 0.996c \).