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A proton \(\left(\mathrm{m}=1.67 \times 10^{-27} \mathrm{~kg}\right)\) is accelerated to a kinetic energy of \(200 \mathrm{MeV}\). What is its speed at this energy?

Short Answer

Expert verified
The proton's speed is approximately \(2.63 \times 10^8 \, \text{m/s}\).

Step by step solution

01

Convert Kinetic Energy to Joules

The given kinetic energy is in mega-electronvolts (MeV). First, we need to convert it to joules. We know that 1 MeV is equivalent to \(1.60218 \times 10^{-13}\) joules. Thus, for 200 MeV, the energy in joules is:\[E_k = 200 \, \text{MeV} \times 1.60218 \times 10^{-13} \, \text{J/MeV} = 3.20436 \times 10^{-11} \, \text{J}\]
02

Use the Relativistic Kinetic Energy Formula

The kinetic energy for a relativistic particle is given by:\[E_k = (\gamma - 1)m c^2\]where \( \gamma \) is the Lorentz factor and \( c \) is the speed of light. Rearranging for \( \gamma \), we have:\[\gamma = 1 + \frac{E_k}{m c^2}\]
03

Calculate Lorentz Factor \(\gamma\)

Substitute the given values into the Lorentz factor formula. The speed of light \( c \) is \(3.00 \times 10^8 \, \text{m/s}\) and the proton mass \( m \) is \(1.67 \times 10^{-27} \, \text{kg}\):\[\gamma = 1 + \frac{3.20436 \times 10^{-11} \, \text{J}}{1.67 \times 10^{-27} \, \text{kg} \times (3.00 \times 10^8 \, \text{m/s})^2}\]Calculate:\[\gamma \approx 1.213\]
04

Solve for Speed using \(\gamma\)

The Lorentz factor is also related to the speed \( v \) by the equation:\[\gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\]Rearranging gives:\[\left(\frac{v}{c}\right)^2 = 1 - \frac{1}{\gamma^2}\]Solving for \( v \), we find:\[\v = c \sqrt{1 - \frac{1}{\gamma^2}}\]Substitute \( \gamma = 1.213 \) into the equation:\[v = 3.00 \times 10^8 \, \text{m/s} \times \sqrt{1 - \frac{1}{1.213^2}}\]Calculate:\[v \approx 2.63 \times 10^8 \, \text{m/s}\]
05

Conclusion

The final speed of the proton after being accelerated to a kinetic energy of 200 MeV is approximately \(2.63 \times 10^8 \, \text{m/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lorentz factor
In relativistic physics, the Lorentz factor (\(\gamma\)) plays a crucial role when dealing with objects moving at speeds comparable to the speed of light. This factor can be thought of as a measure of how much time, length, and relativistic mass change for an object moving close to light speed.
It is defined through the equation:\[\gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\]
  • \(v\) represents the speed of the object.
  • \(c\) is the speed of light, approximately \(3.00 \times 10^8 \, \text{m/s}\).

As an object's speed approaches the speed of light, the denominator of the equation above approaches zero, causing \(\gamma\) to increase drastically. This effectively means that the relativistic effects become more pronounced. In the context of this exercise, by calculating the Lorentz factor, we can understand how the proton behaves when it possesses a large amount of kinetic energy that's significant compared to its rest energy.
Kinetic energy conversion
When dealing with particles in relativistic speeds, like a proton accelerated to high energies, we need to take extra care in converting kinetic energy from one unit to another and using it appropriately in formulas. Initially, the kinetic energy is given in mega-electronvolts (MeV), which is common in particle physics. Conversion to joules is done using:\[1 \, \mathrm{MeV} = 1.60218 \times 10^{-13} \, \mathrm{J}\]So for 200 MeV, this becomes:\[E_k = 200 \, \mathrm{MeV} \times 1.60218 \times 10^{-13} \, \mathrm{J/MeV} = 3.20436 \times 10^{-11} \, \mathrm{J}\]
It’s critical to convert energies into consistent units (like joules here) to correctly compute subsequent quantities and factors, such as the Lorentz factor, which uses mass in kilograms and speed in meters per second. This ensures that our calculations for speed, energy, and other related values remain precise and reliable.
Speed of light
The speed of light, denoted by \(c\), is a core constant in physics with a value of \(3.00 \times 10^8 \, \text{m/s}\). It represents the ultimate speed limit in the universe, as no object with mass can reach or exceed this speed. In relativistic kinematics, which handles objects moving at or near this speed, \(c\) is critical as a conversion factor.
This speed comes up frequently, such as when calculating the Lorentz factor or when determining relativistic kinetic energy and momentum. Particularly in the exercise at hand, we use \(c\) to express the proton's speed as a fraction of the speed of light, helping us understand how relativity affects the particle:\[v = c \sqrt{1 - \frac{1}{\gamma^2}}\]Here, \(v\) is the speed as obtained after computing the Lorentz factor, indicating how close the speed of the proton is to \(c\), thus revealing the significant relativistic effects exerted on the proton when accelerating near the speed of light.

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Most popular questions from this chapter

The clock on a spaceship shows that a robot on board took \(10.0 \mathrm{~s}\) to do some job. The ship flies passed a station, and someone watching the robot also notes how long it took to do the job using her own wristwatch. If she computes \(y\) for the ship to be \(1.08\), how long will she say the robot took to do the job? [Hint: Time dilation means time slows and durations appear longer.]

Compute the rest energy of an electron-that is, the energy equivalent of its mass, \(9.11 \times 10^{-31} \mathrm{~kg}\).

Find the speed and momentum of a proton \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) that has been accelerated through a potential difference of 2000 MV. (We call this a \(2 \mathrm{GeV}\) proton.) Give your answers to three significant figures.

Two cells that subdivide on Earth every \(10.0 \mathrm{~s}\) start from the Earth on a journey to the Sun \(\left(1.50 \times 10^{11} \mathrm{~m}\right.\) away) in \(\mathrm{a}\) spacecraft moving at \(0.850 \mathrm{c}\). How many cells will exist when the spacecraft crashes into the Sun? According to Earth observers, with respect to whom the cells are moving, the time taken for the trip to the Sun is the distance traveled \((x)\) over the speed \((v)\) $$ \Delta t_{\mathrm{M}}=\frac{x}{v}=\frac{1.50 \times 10^{11} \mathrm{~m}}{(0.850)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}=588 \mathrm{~s} $$ Because spacecraft clocks are moving with respect to the planet, they appear from Earth to run more slowly. The time these clocks read is $$ \Delta t_{\mathrm{s}}=\Delta f_{\mathrm{M}} / \gamma=\Delta t_{\mathrm{M}} \sqrt{1-(v / \mathrm{c})^{2}} $$ and so $$ \Delta l_{\mathrm{s}}=310 \mathrm{~s} $$ The cells divide according to the spacecraft clock, a clock that is at rest relative to them. They therefore undergo 31 divisions in this time, since they divide each \(10.0 \mathrm{~s}\). Therefore, the total number of cells present on crashing is $$ (2)^{31}=2.1 \times 10^{9} \text { cells } $$

A \(2.0\) -kg object is lifted from the floor to a tabletop \(30 \mathrm{~cm}\) above the floor. By how much did the mass of the system consisting of the Earth and the object increase because of this increased \(\mathrm{PE}_{\mathrm{G}}\) ? We use \(\Delta \mathrm{E}_{\mathrm{O}}=(\Delta m) \mathrm{c}^{2}\), with \(\Delta \mathrm{E}_{\mathrm{O}}=m g h .\) Therefore, $$ \Delta m=\frac{\Delta E_{0}}{c^{2}}=\frac{m g h}{c^{2}}=\frac{(2.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.30 \mathrm{~m})}{\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}}=6.5 \times 10^{-17} \mathrm{Kg} $$

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