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?

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.