## Units in Particle Physics: Practice Problems and Answers |

## Practice Problems

- A proton in the Fermilab Tevatron has an energy of 1.00 TeV. What is this energy in Joules?

- A 2.00 g bug moves at 0.100 cm/s. Find its momentum and kinetic energy in TeV.

- The mass of a proton is 1.67 x 10
^{-27}kg. Convert this to GeV and TeV.

- The mass of an electron is 0.511 MeV. Convert to kg.

- The electron accelerator at Jefferson Lab in Virginia runs at a beam energy of about a 5.5 GeV. Because they have so little mass, we can say the momentum of any one of these electrons is also 5.5 GeV. What is that in kg-m/s?

## Solutions

1.609 x 10(see table)^{-7}J

- p = mv = (2.00 x 10
^{-3}kg)(1.00 x 10^{-3}m/s) = 2.00 x 10^{-6}kg-m/s

Then (2.00 x 10^{-6}kg-m/s) / (5.36 x 10^{-16}kg-m-c/s-TeV) =3.73 x 10^{9}TeV/c

K = 0.5mv^{2}= 0.5(2.00 x 10^{-3}kg)1.00 x 10^{-3}m/s)^{2}= 1.00 x 10^{-9}J

Then (1.00 x 10^{-9}J) / (1.609 x 10^{-7}J/TeV) =6.22 x 10^{-3}TeV

- (1.67 x 10
^{-27}kg) / (1.79 x 10^{-27}kg-c^{2}/GeV) =0.933 GeV/c^{2}

The quick way to get TeV is to divide by 1000: (0.933 GeV/c^{2}) / (1000 TeV/GeV) =9.33 x 10^{-4}TeV/c^{2}

Note: The actual proton mass, calculated more precisely, is closer to 0.938 GeV/c^{2}.

- (0.511 MeV/c
^{2})(1.79 x10^{-30}kg-c^{2}/MeV) =9.15 x 10^{-31}kg

- (5.5 GeV/c)(5.36 x 10
^{-19}kg-m-c/s-GeV) =3.0 x 10^{-18}kg-m/s

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Last Update: February 7, 2002