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Chapter 20: Problem 29

Calculate the speed parameter \(\beta\) of a particle with a momentum of \(12.5 \mathrm{MeV} / c\) if the particle is \((a)\) an electron and \((b)\) a proton.

Short Answer

Expert verified
The speed parameter \( \beta \) is calculated to be approximately 0.930 and 0.014 for the electron and proton, respectively.

Step by step solution

01

Identify Given Information and Formulas

Firstly, it's important to gather the necessary parameters. The problem gives that the momentum of the particle, denoted by \( p \), is 12.5 MeV/c. The formula to be used is the one for relativistic momentum (\( p = \gamma m v \)), and the definition of speed parameter (\( \beta = \frac{v}{c} \)).
02

Calculate β for the Electron

Using the relativistic momentum formula and the definition of speed parameter, rewrite p in terms of m and β to get the formula for β. For an electron, the mass (m) is approximately 0.511 MeV/c². Substitute the given momentum and the mass of the electron into the equation and solve for β.
03

Calculate β for the Proton

For a proton, the mass (m) is approximately 938 MeV/c². Just like with the electron, substitute these values into the equation derived in step 2 and solve for β. Using a calculator, make sure to apply the correct order of operations (multiplication and division before addition and subtraction).

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