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Chapter 26: Problem 77

A charged particle is observed to have a total energy of \(0.638 \mathrm{MeV}\) when it is moving at \(0.600 c .\) If this particle enters a linear accelerator and its speed is increased to \(0.980 c,\) what is the new value of the particle's total energy?

Short Answer

Expert verified
Answer: To find the new total energy of the particle, follow these steps: 1. Write down the relativistic energy-momentum equation: \(E^2 = (mc^2)^2 + (\gamma mvc)^2\) 2. Write the expression for the relativistic momentum: \(p = \gamma mv\), where \(\gamma=\frac{1}{\sqrt{1-v^2/c^2}}\). 3. Use the given initial total energy (0.638 MeV) and initial speed (0.6c) to solve for the rest mass energy (mc^2). 4. Calculate the new Lorentz factor \(\gamma'\) for the new speed (0.980c): \(\gamma' = \frac{1}{\sqrt{1-v'^2/c^2}}\) 5. Use the new Lorentz factor \(\gamma'\) and the rest mass energy (mc^2) to calculate the new total energy E': \(E'^2 = (mc^2)^2 + (\gamma' mvc)^2\) After solving for E', you will have the new total energy of the charged particle after it passes through the linear accelerator.

Step by step solution

01

Write down the relativistic energy-momentum equation

The relativistic energy-momentum equation is given by: \(E^2 = (mc^2)^2 + (pc)^2\) where E is the total energy, m is the rest mass, c is the speed of light, and p is the relativistic momentum.
02

Write the expression for the relativistic momentum

The relativistic momentum (p) is given by: \(p = \gamma mv\) where \(\gamma=\frac{1}{\sqrt{1-v^2/c^2}}\) is the Lorentz factor, m is the rest mass, and v is the speed of the particle. Thus, the relativistic energy-momentum equation becomes: \(E^2 = (mc^2)^2 + (\gamma mvc)^2\)
03

Solve for the rest mass energy (mc^2)

Using the given initial total energy E, and the initial speed (v) of the charged particle, solve the energy-momentum equation for the rest mass energy (mc^2). \(E = 0.638 MeV = 0.638 \times 10^6 eV\) \(v = 0.6c\) Plug these values into the energy-momentum equation: \(E^2 = (mc^2)^2 + (\gamma mvc)^2\) \((0.638 \times 10^6 eV)^2 = (mc^2)^2 + (\frac{0.6mc^2}{\sqrt{1-(0.6)^2}})^2\) Now solve for (mc^2), the rest mass energy.
04

Calculate the new Lorentz factor

The new speed of the particle is given by \(v' = 0.980c\). Calculate its new Lorentz factor \(\gamma'\) by using the formula: \(\gamma' = \frac{1}{\sqrt{1-v'^2/c^2}} = \frac{1}{\sqrt{1-(0.980)^2}}\)
05

Calculate the new total energy E'

Now, use the new Lorentz factor \(\gamma'\), and the rest mass energy (mc^2) that we found in step 3 to calculate the new total energy E': \(E'^2 = (mc^2)^2 + (\gamma' mvc)^2\) Plug in the values of mc^2 and \(\gamma'\) and solve for E'. Convert the result back to MeV. This will give us the new value of the charged particle's total energy after it passes through the linear accelerator.

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Most popular questions from this chapter

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