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Chapter 37: Problem 62

In certain radioactive beta decay processes, the beta particle (an electron) leaves the atomic nucleus with a speed of 99.95\(\%\) the speed of light relative to the decaying nucleus. If this nucleus is moving at 75.00\(\%\) the speed of light in the laboratory reference frame, find the speed of the emitted electron relative to the laboratory reference frame if the electron is emitted (a) in the same direction that the nucleus is moving and (b) in the opposite direction from the nucleus's velocity. (c) In each case in parts (a) and (b), find (b), find the kinetic energy of the electron as measured in (i) the laboratory frame and (ii) the reference frame of the decaying nucleus.

Short Answer

Expert verified
(a) 0.999428c; (b) -0.999500c; KE in lab: (a) 36.77mc^2, (b) 49mc^2; KE in nucleus: 49mc^2 for both.

Step by step solution

01

Define Variables

Let \( v_n = 0.75c \) be the velocity of the nucleus relative to the laboratory and \( v_e' = 0.9995c \) be the velocity of the electron relative to the nucleus. Our goal is to find the velocity of the electron \( v_e \) relative to the laboratory, where \( c \) is the speed of light.
02

Use the Relativistic Velocity Addition Formula

We use the relativistic addition of velocities formula: \( v_e = \frac{v_n + v_e'}{1 + \frac{v_n v_e'}{c^2}} \) for part (a), where both are in the same direction. Substitute \( v_n = 0.75c \) and \( v_e' = 0.9995c \).
03

Calculate the Electron Velocity in Part (a)

For part (a):\[ v_{ea} = \frac{0.75c + 0.9995c}{1 + \frac{0.75 \times 0.9995}{1}} = \frac{1.7495c}{1.749625} \approx 0.999428c \]The electron moves at approximately 0.999428c relative to the laboratory.
04

Use the Relativistic Velocity Addition Formula (Opposite Direction)

For part (b), the electron and nucleus move in opposite directions. Use:\[ v_{e} = \frac{v_n - v_e'}{1 - \frac{v_n v_e'}{c^2}} \] Substitute \( v_n = 0.75c \) and \( v_e' = 0.9995c \).
05

Calculate the Electron Velocity in Part (b)

For part (b):\[ v_{eb} = \frac{0.75c - 0.9995c}{1 - \frac{0.75 \times 0.9995}{1}} = \frac{-0.2495c}{0.249625} \approx -0.999500c \] The electron moves at approximately -0.999500c relative to the laboratory (the negative sign indicates opposite direction).
06

Calculate the Kinetic Energy in the Laboratory Frame (Part a)

The relativistic kinetic energy is given by \( KE = (\gamma - 1)mc^2 \) where \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \). For part (a), electron velocity \( v = 0.999428c \):\[ \gamma_a = \frac{1}{\sqrt{1 - 0.999428^2}} \approx 37.77 \]\[ KE_{L,a} = (37.77 - 1)mc^2 = 36.77mc^2 \]
07

Calculate the Kinetic Energy in the Laboratory Frame (Part b)

For part (b), using \( v = 0.999500c \):\[ \gamma_b = \frac{1}{\sqrt{1 - 0.999500^2}} \approx 50 \]\[ KE_{L,b} = (50 - 1)mc^2 = 49mc^2 \]
08

Calculate the Kinetic Energy in the Nucleus Frame

In the nucleus frame, the kinetic energy for both parts (a) and (b) is:\( v = 0.9995c \):\[ \gamma' = \frac{1}{\sqrt{1 - 0.9995^2}} \approx 50 \]\[ KE' = (50 - 1)mc^2 = 49mc^2 \]

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Key Concepts

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

Beta Decay
Beta decay is a fascinating process observed in certain unstable atomic nuclei. During this process, a neutron in the nucleus is transformed into a proton, emitting a beta particle, which is essentially an electron or a positron, along with an antineutrino or neutrino.
Beta decay occurs due to the weak nuclear force, one of the four fundamental forces in nature. This decay changes the atomic number of the element, leading to the formation of a different element.
  • The beta particle (electron or positron) is emitted with significant energy, often reaching speeds close to the speed of light.
  • This emitted particle can either increase or decrease the atomic number of the nucleus depending on whether an electron or positron is released.
When solving problems related to beta decay, it's crucial to consider its impact on atomic identity and how the high-speed particles are modeled through special relativity.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. In classical physics, it's simply calculated using the formula \[ KE = \frac{1}{2}mv^2 \] where \(m\) is mass and \(v\) is velocity.
However, when dealing with particles moving at speeds close to the speed of light, such as in beta decay, we need to apply relativistic physics. Here, kinetic energy is calculated using: \[ KE = (\gamma - 1)mc^2 \] where \[ \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \] is the Lorentz factor, representing time dilation and length contraction effects.
  • Relativistic kinetic energy accounts for the fact that as a particle鈥檚 speed approaches the speed of light, its energy increases dramatically, and it never reaches \(c\) due to its infinite energy requirement at that point.
  • This formula is crucial for calculating the energy changes in high-speed processes like beta decay, providing accurate predictions of the emitted particle's energy.
Special Relativity
Special relativity is a theory formulated by Albert Einstein that revolutionized our understanding of space, time, and energy. It is especially important in physics problems involving high velocities.
One key aspect of the theory is how it handles the addition of velocities, known as relativistic velocity addition. This comes into play when considering the velocities of beta particles emitted during decay.In a relativistic context, we're interested in how velocities combine differently than in classical mechanics. The formula is:
  • For particles moving in the same direction: \[ v_{\text{relative}} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} \]
  • For opposite directions: \[ v_{\text{relative}} = \frac{v_1 - v_2}{1 - \frac{v_1 v_2}{c^2}} \]
These formulas ensure that even at extreme speeds, results remain consistent with the speed of light being the ultimate speed limit in our universe.
Atomic Physics
Atomic physics explores the fundamental components and interactions within atoms. Understanding beta decay and related phenomena requires a solid foundation in this field of physics.
Atoms consist of a nucleus made up of protons and neutrons, surrounded by electrons. In beta decay, changes within the nucleus alter the identity and energy of atoms. Key principles include:
  • Protons define the element. A change in atomic number changes the element identity.
  • Neutrons contribute to atomic mass and stability but not chemical identity directly.
Atomic physics also studies electron configurations and transitions, which are crucial in comprehending how particles like beta particles are emitted and how they interact with their surroundings. This understanding enables accurate modeling and prediction of behaviors in high-energy events, like beta decay.

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Most popular questions from this chapter

A spacecraft flies away from the earth with a speed of \(4.80 \times 10^{6} \mathrm{m} / \mathrm{s}\) relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days \((1\) year) later, as measured by the clock that remained on earth. What is the difference in the elapsed times on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on earth, shows the shorter elapsed time?

The negative pion \(\left(\pi^{-}\right)\) is an unstable particle with an average lifetime of \(2.60 \times 10^{-8}\) s (measured in the rest frame of the pion). (a) If the pion is made to travel at very high speed relative to a laboratory, its average lifetime is measured in the laboratory to be \(4.20 \times 10^{-7} \mathrm{s}\) . Calculate the speed of the pion expressed as a fraction of \(c .\) (b) What distance, measured in the laboratory, does the pion travel during its average lifetime?

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c,\) what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

Inside a spaceship flying past the earth at three-fourths the speed of light, a pendulum is swinging. (a) If each swing takes 1.50 s as measured by an astronaut performing an experiment inside the spaceship, how long will the swing take as measured by a person at mission control on earth who is watching the experiment? (b) If each swing takes 1.50 s as measured by a person at mission control on earth, how long will it take as measured by the astronaut in the spaceship?

(a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-\) vt and \(t^{\prime}=t .\) In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $$\frac{\partial^{2} E(x, t)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E(x, t)}{\partial t^{2}}=0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S^{\prime}\) is found to be $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ This has a different form than the wave equation in \(S .\) Hence the Galilean transformation violates the first relativity postulate that all physical laws have the same form in all inertial reference frames. (Hint: Express the derivatives \(\partial / \partial x\) and \(\partial / \partial t\) in terms of \(\partial / \partial x^{\prime}\) and \(\partial / \partial t^{\prime}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Egs. \((37.21),\) and show that in frame \(S^{\prime}\) the wave equation has the same form as in frame \(S :\) $$\frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ Explain why this shows that the speed of light in vacuum is \(c\) in both frames \(S\) and \(S^{\prime} .\)
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