Re: How fast does a 30 keV electron travel?

Date: Wed Oct 4 11:11:16 2000
Posted By: Samuel Silverstein, faculty, physics, Stockholm University
Area of science: Physics
ID: 969812437.Ph

Message:

When we talk about a 30 keV (thousand electron-volts) electron, the 30 keV refers to its relativistic momentum P, defined as

P = cp

where p is the momentum and c is the speed of light. The total energy includes the electron's rest mass M as well, defined as:

    						
M = mc^2 = 511 keV

The total energy E of this 30 keV electron is calculated as:

E^2 = M^2 + P^2

Substituting M=511 keV and P=30 KeV, we get E=511.8799 or so. As you can see, the total energy of the electron is still dominated by the rest mass, so we can guess we aren't that close to the speed of light.

Now, Einstein tells us that the total energy E can be expressed as:

          mc^2   
E = -----------------
    sqrt(1 - (v/c)^2)

We already know mc^2 equals 511 keV, so we can solve for v/c. The final result is:

v/c = 0.0586

meaning that the electron is travelling slightly less than 6 percent of light speed (about 17.6 million meters/second).

This may not have been the fastest way to do the calculation, but I hope it was straightforward enough for you to follow the logic, and do further calculations of your own.

Regards,
Sam Silverstein

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