# Proton’s kinetic energy

## Example: Proton’s kinetic energy

A proton (m = 1.67 x 10-27 kg) travels at a speed v = 0.9900c = 2.968 x 108m/s. What is its kinetic energy?

According to a classical calculation, which is not correct, we would obtain:

K = 1/2mv2 = ½ x (1.67 x 10-27 kg) x (2.968 x 108m/s)2 = 7.355 x 10-11 J

With relativistic correction the relativistic kinetic energy is equal to:

K = (ɣ – 1)mc2

where the Lorentz factor

ɣ = 7.089

therefore

K = 6.089 x (1.67 x 10-27 kg) x (2.9979 x 108m/s)2 = 9.139 x 10-10 J = 5.701 GeV

This is about 12 times higher energy as in the classical calculation. According to this relationship, an acceleration of a proton beam to 5.7 GeV requires energies that are in the order different.

References:
Reactor Physics and Thermal Hydraulics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. Todreas Neil E., Kazimi Mujid S. Nuclear Systems Volume I: Thermal Hydraulic Fundamentals, Second Edition. CRC Press; 2 edition, 2012, ISBN: 978-0415802871
6. Zohuri B., McDaniel P. Thermodynamics in Nuclear Power Plant Systems. Springer; 2015, ISBN: 978-3-319-13419-2
7. Moran Michal J., Shapiro Howard N. Fundamentals of Engineering Thermodynamics, Fifth Edition, John Wiley & Sons, 2006, ISBN: 978-0-470-03037-0
8. Kleinstreuer C. Modern Fluid Dynamics. Springer, 2010, ISBN 978-1-4020-8670-0.
9. U.S. Department of Energy, THERMODYNAMICS, HEAT TRANSFER, AND FLUID FLOW. DOE Fundamentals Handbook, Volume 1, 2 and 3. June 1992.

Kinetic Energy