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The speed of a fast moving electron, having total energy of 2 MeV, is nearly:
- a)0.96 c
- b)0.99 c
- c)0.98 c
- d)0.97 c
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The speed of a fast moving electron, having total energy of 2 MeV, is ...
Given particle is electron, the mass of electron is m = 9.11 × 10-31 kg, c = 3 × 108 m/s
1 eV = 1.602 × 10-19 J
E = 2 MeV = 2 × 106 × 1.602 × 10-19 J = 3.204 × 10-13 J
Now from the above formula:
Most Upvoted Answer
The speed of a fast moving electron, having total energy of 2 MeV, is ...
Understanding the Problem
To determine the speed of a fast-moving electron with a total energy of 2 MeV, we need to apply some concepts from relativistic physics.
Key Concepts
- Total Energy (E): In relativistic terms, the total energy of a particle can be expressed as:
E = K + mc²
where K is the kinetic energy and mc² is the rest energy of the particle.
- Rest Mass Energy of Electron: The rest mass energy (mc²) of an electron is approximately 0.511 MeV.
Calculation Steps
1. Calculate Kinetic Energy (K):
- Given total energy E = 2 MeV,
- Rest mass energy mc² = 0.511 MeV,
- Therefore, K = E - mc² = 2 MeV - 0.511 MeV = 1.489 MeV.
2. Relativistic Relationship:
- The relation between speed (v), kinetic energy (K), and total energy (E) is given by:
\[
E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}
\]
3. Solving for Speed (v):
- Rearranging the equation, we find:
\[
v = c \sqrt{1 - \left( \frac{mc^2}{E} \right)^2}
\]
4. Substituting Values:
- Substitute mc² = 0.511 MeV and E = 2 MeV into the equation.
5. Final Calculation:
- This results in:
\[
v \approx c \sqrt{1 - (0.511/2)^2} \approx 0.97c
\]
Conclusion
Thus, the speed of the electron is approximately 0.97c, which corresponds to option 'D'. This high speed indicates that the electron is moving relativistically, demonstrating significant kinetic energy compared to its rest mass energy.
To determine the speed of a fast-moving electron with a total energy of 2 MeV, we need to apply some concepts from relativistic physics.
Key Concepts
- Total Energy (E): In relativistic terms, the total energy of a particle can be expressed as:
E = K + mc²
where K is the kinetic energy and mc² is the rest energy of the particle.
- Rest Mass Energy of Electron: The rest mass energy (mc²) of an electron is approximately 0.511 MeV.
Calculation Steps
1. Calculate Kinetic Energy (K):
- Given total energy E = 2 MeV,
- Rest mass energy mc² = 0.511 MeV,
- Therefore, K = E - mc² = 2 MeV - 0.511 MeV = 1.489 MeV.
2. Relativistic Relationship:
- The relation between speed (v), kinetic energy (K), and total energy (E) is given by:
\[
E = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}
\]
3. Solving for Speed (v):
- Rearranging the equation, we find:
\[
v = c \sqrt{1 - \left( \frac{mc^2}{E} \right)^2}
\]
4. Substituting Values:
- Substitute mc² = 0.511 MeV and E = 2 MeV into the equation.
5. Final Calculation:
- This results in:
\[
v \approx c \sqrt{1 - (0.511/2)^2} \approx 0.97c
\]
Conclusion
Thus, the speed of the electron is approximately 0.97c, which corresponds to option 'D'. This high speed indicates that the electron is moving relativistically, demonstrating significant kinetic energy compared to its rest mass energy.
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The speed of a fast moving electron, having total energy of 2 MeV, is nearly:a)0.96 cb)0.99 cc)0.98 cd)0.97 cCorrect answer is option 'D'. Can you explain this answer?
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