/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 An electron has a speed of \(0.7... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron has a speed of \(0.75 c\). Find the speed of a proton that has (a) the same kinetic energy as the electron and (b) the same momentum as the electron.

Short Answer

Expert verified
The speed of a proton that has (a) the same kinetic energy as the electron and (b) the same momentum as the electron can be calculated by rearranging and solving the formulas for kinetic energy and momentum respectively.

Step by step solution

01

Calculate initial kinetic energy of the electron

Firstly, calculate the kinetic energy of the electron using the formula \( KE = \frac{{m_0 c^2}}{{\sqrt{1 - v^2/c^2}}} - m_0 c^2 \) where \( KE \) is the kinetic energy, \( m_0 \) is the rest mass of the electron which is 9.11 * 10^-31 kg, \( c \) is the speed of light (3 * 10^8 m/s) and \( v \) is the velocity of electron which is already given as 0.75c.
02

Apply kinetic energy to proton for Part (a)

Use the kinetic energy of electron calculated in the previous step to find the speed of the proton (rest mass 1.67 * 10^-27kg) which has the same kinetic energy. Solve the equation obtained in Step 1 for \( v \) to get the speed of proton.
03

Calculate initial momentum of the electron

The classical momentum equation is not valid in this case due to high speeds. Therefore, we use the equation \( p=\frac{{m_0 v}}{{\sqrt{1 - v^2/c^2}}}\) to calculate momentum, where \( p \) is the momentum.
04

Apply momentum to proton for Part (b)

Use the momentum of electron calculated in step 3 to find the speed of the proton which has the same momentum. Solve the equation obtained in step 3 for \( v \) to get the speed of proton.

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Key Concepts

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

Kinetic Energy
In relativistic mechanics, kinetic energy takes on a new form compared to classical mechanics. While in classical physics kinetic energy is calculated as \( KE = \frac{1}{2}mv^2 \), in relativistic physics it incorporates the speed of light and the rest mass of the object. This is because the effects of relativity become significant at speeds close to the speed of light. The formula used is \( KE = \frac{{m_0 c^2}}{{\sqrt{1 - v^2/c^2}}} - m_0 c^2 \), which accounts for the increase in mass and energy as objects move faster.
  • Rest mass \( m_0 \) is the mass of an object when it is at rest.
  • \( c \) is the speed of light, an essential constant in all relativistic equations.
To find the kinetic energy of an electron moving at \( 0.75c \), substitute its rest mass and speed into the formula. Then, using this kinetic energy, you can solve for the speed of a proton that has the same kinetic energy by equating the two energies and solving for the proton's velocity.
Momentum
Momentum in relativistic mechanics requires a different approach compared to its classical counterpart. Although classically calculated using \( p = mv \), relativistic momentum must consider the speed of light, especially at velocities nearing this limit. The formula evolves to \( p=\frac{{m_0 v}}{{\sqrt{1 - v^2/c^2}}} \). This equation ensures momentum approaches infinity as an object's speed approaches the speed of light, reflecting the harder it becomes to accelerate the object further.
  • An object's momentum will increase without bound as its speed nears the speed of light.
  • Relativistic momentum is crucial in high-speed particle physics.
When given an electron's speed of \( 0.75c \), you compute its momentum with this formula. Next, you can calculate the speed of a proton, having equivalent momentum, by equaling the proton's relativistic momentum to that of the electron and solving for its velocity.
Speed of Light
The speed of light is a fundamental constant in physics, denoted as \( c \), approximately equal to \( 3 \times 10^8 \) meters per second. It serves as a crucial limit in relativity theory. No object with mass can attain or exceed this speed as it would require infinite energy.
  • Light speed is the highest possible speed for any physical entity.
  • Relativity fundamentally changes our understanding of space-time at speeds close to \( c \).
In problems involving relativistic speeds, like the given electron traveling at \( 0.75c \), using the speed of light within equations allows accurate calculation of kinetic energy and momentum. Techniques accounting for \( c \) enable precise predictions of behavior for particles traveling at significant fractions of light speed, essential for understanding high-energy physics and the universe's fundamental laws.

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Most popular questions from this chapter

Electrons in projection television sets are accelerated through a total potential difference of \(50,000 \mathrm{~V}\). (a) Calculate the speed of the electrons using the relativistic form of kinetic energy assuming the electrons start from rest. (b) Calculate the speed of the electrons using the classical form of kinetic energy. (c) Is the difference in speed significant in the design of this set in your opinion?

A radium isotope decays to a radon isotope, \({ }^{222} \mathrm{Rn}\), by emitting an \(\alpha\) particle (a helium nucleus) according to the decay scheme \({ }^{226} \mathrm{Ra} \rightarrow{ }^{222} \mathrm{Rn}+{ }^{4} \mathrm{He}\). The masses of the atoms are \(226.0254\) (Ra), \(222.0175\) (Rn), and \(4.0026\) (He). How much energy is released as the result of this decay?

The creation and study of new elementary particles is an important part of contemporary physics. Especially interesting is the discovery of a very massive particle. To create a particle of mass \(M\) requires an energy \(M c^{2}\). With enough energy, an exotic particle can be created by allowing a fast-moving particle of ordinary matter, such as a proton, to collide with a similar target particle. Let us consider a perfectly inelastic collision between two protons: An incident proton with mass \(m\), kinetic energy \(K\), and momentum magnitude \(p\) joins with an originally stationary target proton to form a single product particle of mass \(M .\) You might think that the creation of a new product particle, 9 times more massive than in a previous experiment, would require just 9 times more energy for the incident proton. Unfortunately, not all of the kinetic energy of the incoming proton is available to create the product particle, since conservation of momentum requires that after the collision the system as a whole still must have some kinetic energy. Only a fraction of the energy of the incident particle is thus available to create a new particle. You will determine how the energy available for particle creation depends on the energy of the moving proton. Show that the energy available to create a product particle is given by $$ M c^{2}=2 m c^{2} \sqrt{1+\frac{K}{2 m c^{2}}} $$ From this result, when the kinetic energy \(K\) of the incident proton is large compared to its rest energy \(m c^{2}\), we see that \(M\) approaches \((2 m K)^{1 / 2} / c .\) Thus if the energy of the incoming proton is increased by a factor of 9, the mass you can create increases only by a factor of \(3 .\) This disappointing result is the main reason that most modern accelerators, such as those at CERN (in Europe), at Fermilab (near Chicago), at SLAC (at Stanford), and at DESY (in Germany), use colliding beams. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so in principle all of the initial kinetic energy can be used for particle creation, according to $$ M c^{2}=2 m c^{2}+K=2 m c^{2}\left(1+\frac{K}{2 m c^{2}}\right) $$ where \(K\) is the total kinetic energy of two identical colliding particles. Here, if \(K>>m c^{2}\), we have \(M\) directly proportional to \(K\), as we would desire. These machines are difficult to build and to operate, but they open new vistas in physics.

As noted in Section \(2.2\), the quantity \(E-p^{2} c^{2}\) is an \(i n-\) variant in relativity theory. This means that the quantity \(E^{2}-p^{2} c^{2}\) has the same value in all inertial frames even though \(E\) and \(p\) have different values in different frames. Show this explicitly by considering the following case. A particle of mass \(m\) is moving in the \(+x\) direction with speed \(u\) and has momentum \(p\) and energy \(E\) in the frame \(S\). (a) If \(S^{\prime}\) is moving at speed \(v\) in the standard way, find the momentum \(p^{\prime}\) and energy \(E^{\prime}\) observed in \(\mathrm{S}^{\prime} .\) (Hint: Use the Lorentz velocity transformation to find \(p^{\prime}\) and \(E^{\prime} .\) Does \(E=E^{\prime}\) and \(p=p^{\prime} ?\) (b) Show that \(E^{2}-p^{2} c^{2}\) is equal to \(E^{\prime 2}-p^{\prime 2} c^{2}\).

A charged particle moves along a straight line in a uniform electric field \(E\) with a speed \(v\). If the motion and the electric field are both in the \(x\) direction, (a) show that the magnitude of the acceleration of the charge \(q\) is given by $$ a=\frac{d v}{d t}=\frac{q E}{m}\left(1-\frac{v^{2}}{c^{2}}\right)^{3 / 2} $$ (b) Discuss the significance of the dependence of the acceleration on the speed. (c) If the particle starts from rest at \(x=0\) at \(t=0\), find the speed of the particle and its position after a time \(t\) has elapsed. Comment on the limiting values of \(v\) and \(x\) as \(t \rightarrow \infty\)

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