For part c), how come we cannot use the relativistic momentum equation where p = mv/(sqrt(1-v^2/c^2))? Since we've already solved for velocity in part b)? Imagine you are able to observe a 7.5 MeV electron in the lab. Using the equationsrelating to relativistic energy and momentum, answer the following questions.a. What is the total energy of the electron?E÷KE + mc² = 2×108 + 7.5X106 = 2.075 × 108 Vb. What is the velocity of the electron?mc2honV2сE² = [mc²)² + c²P=E²-(mc²) ²✓ =C. What is the momentum of the electron?C2√ in- 2.996×10-0.83 eVs/mm² cuего8m/s

Kinetic energy KE = 7.5MeV rest mass energy of electron U = 0.51MeV

Answered: Imagine you are able to observe a 7.5…

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter31: Radioactivity And Nuclear Physics

Section: Chapter Questions

Problem 16PE: (a) What is the kinetic energy in MeV of a ray that is traveling at 0.998c? This gives some idea of...

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For part c), how come we cannot use the relativistic momentum equation where p = mv/(sqrt(1-v^2/c^2))? Since we've already solved for velocity in part b)?

Imagine you are able to observe a 7.5 MeV electron in the lab. Using the equations
relating to relativistic energy and momentum, answer the following questions.
a. What is the total energy of the electron?
E÷KE + mc² = 2×108 + 7.5X106 = 2.075 × 108 V
b. What is the velocity of the electron?
mc2
hon
V2
с
E² = [mc²)² + c²
P=E²-(mc²) ²
✓ =
C. What is the momentum of the electron?
C2
√ in
- 2.996×10
-0.83 eVs/m
m² cu
его
8
m/s

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