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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 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) Electron speed: 0.999875c; (b) Electron speed: -0.42857c. Kinetic energies differ based on direction and frame.

Step by step solution

01

Understand the Problem

We need to calculate the speed and kinetic energy of an electron emitted from a decaying nucleus moving at partial light speed, considering relativistic effects. Both the speed of the nucleus and the speed of the electron are given as fractions of the speed of light (c).
02

Relativistic Velocity Addition - Same Direction

Use the formula for relativistic velocity addition: \[v = \frac{u + v'}{1 + \frac{uv'}{c^2}}\]where \(u = 0.75c\) is the speed of the nucleus, \(v' = 0.9995c\) is the speed of the electron relative to the nucleus, and \(v\) is the electron's speed relative to the lab frame. Calculate \(v\) for the electron emitted in the same direction.
03

Calculate Relative Speed - Same Direction

Substitute the values into the formula:\[v = \frac{0.75c + 0.9995c}{1 + \frac{0.75 \times 0.9995}{1}} \approx 0.999875c\]This is the speed of the electron relative to the lab frame when emitted in the same direction as the nucleus.
04

Relativistic Velocity Addition - Opposite Direction

For the opposite direction, use the same formula:\[v = \frac{u - v'}{1 - \frac{uv'}{c^2}}\]where \(u = 0.75c\) and \(v' = 0.9995c\). Calculate \(v\) for the electron emitted in the opposite direction of the nucleus.
05

Calculate Relative Speed - Opposite Direction

Substitute the values into the formula:\[v = \frac{0.75c - 0.9995c}{1 - \frac{0.75 \times 0.9995}{1}} \approx -0.42857c\]This is the speed of the electron relative to the lab frame when emitted in the opposite direction to the nucleus.
06

Calculate Kinetic Energy in Lab Frame - Same Direction

The relativistic kinetic energy formula is:\[K = (\gamma - 1)mc^2\]where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) and \(v = 0.999875c\). Compute \(K\) using \(m\), the rest mass of the electron (\(9.11 \times 10^{-31}\) kg).
07

Calculate Kinetic Energy in Lab Frame - Opposite Direction

Use the kinetic energy formula for the opposite direction:\[K = (\gamma - 1)mc^2\]where \(v = 0.42857c\). Use relevant calculations to find \(K\).
08

Calculate Kinetic Energy in the Nucleus Frame

For both directions, the kinetic energy of the electron in the nucleus frame (\(v' = 0.9995c\)) is:\[K = (\gamma' - 1)mc^2\]Compute \(K\) using \(v'\) for both cases.

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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 type of radioactive decay where a nucleus transforms by emitting a beta particle, which is typically an electron or positron. This process alters the atomic number of the nucleus, changing one type of element into another. In beta minus decay, a neutron converts into a proton, emitting an electron (the beta particle) and an antineutrino. This particle leaves the nucleus at high speeds, often approaching the speed of light, especially in heavy nuclei.
  • Beta decay has significant implications in nuclear physics and helps explain the transition of elements in nature.
  • This process is a result of weak interaction, one of the four fundamental forces in physics responsible for certain types of nuclear reactions.
In our exercise, the focus is on understanding how the movement of the beta particle during its high-speed ejection affects its observed speed and kinetic energy, especially when relativistic effects are considered.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For objects moving at speeds close to the speed of light, as often seen with particles in nuclear reactions, classical calculations of kinetic energy are insufficient. Instead, relativistic kinetic energy must be calculated using:
\[ K = (\gamma - 1)mc^2 \]where \( \gamma \) (gamma) is the Lorentz factor, defined as:
\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]and \( m \) is the rest mass of the particle.
  • In our context, it is crucial for understanding the enormous energy that high-speed electrons (results of beta decay) attain.
  • This relativistic kinetic energy is substantially higher than what would be predicted by classical mechanics, due to the diminishing effects of increasing speed on velocity.
Calculating kinetic energy in different frames, such as the laboratory versus the nucleus frame, helps depict how motion and energy appear different to observers moving at different speeds.
Relativistic Effects
Relativistic effects become very important when dealing with speeds that are a considerable fraction of the speed of light. Standard Newtonian physics inadequate to explain phenomena because it doesn't account for time dilation or changes in mass. Instead, Einstein's theory of relativity provides the necessary framework.
  • One key effect is time dilation, where time moves slower for a fast-moving object relative to a stationary observer.
  • Length contraction can occur, reducing the perceived length in the direction of motion.
  • Mass appears to increase, which influences relativistic kinetic energy calculations.
In the exercise, the use of relativistic velocity addition shows these principles in action. As electrons are emitted at speeds near the speed of light, relativistic formulas ensure that calculated speeds remain physically possible without exceeding the speed of light. This reveals the limits on speed dictated by relativity and the need to adjust classical views when velocities are high.

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Most popular questions from this chapter

Electromagnetic radiation from a star is observed with an earth-based telescope. The star is moving away from the earth at a speed of 0.520c. If the radiation has a frequency of 8.64 \(\times\) 10\(^{14}\) Hz in the rest frame of the star, what is the frequency measured by an observer on earth?

Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame \(S'\) moving relative to \(S\), the second event is observed to occur 2.15 s after the first. What is the difference between the positions of the two events as measured in \(S'\)?

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when the front of Mavis's ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate x and t as measured by Stanley for the event of turning on the light. (b) Use the time dilation formula, Eq. (37.6), to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Compare to the value of \(t\) you calculated in part (a). (c) Multiply the time interval by Mavis's speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on. Compare to the value of \(x\) you calculated in part (a).

In the earth's rest frame, two protons are moving away from each other at equal speed. In the frame of each proton, the other proton has a speed of 0.700\(c\). What does an observer in the rest frame of the earth measure for the speed of each proton?

(a) Consider the Galilean transformation along the \(x\)-direction : \(x' = x - vt\) and \(t' = t\). In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $${\partial^2E(x, t)\over \partial x^2} - {1 \over c^2} {\partial^2E(x, t)\over \partial t^2} = 0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S'\) is found to be $$(1 - {v^2 \over c^2} ) {\partial^2E(x', t') \over \partial{x'^2}} + {2v \over c^2} {\partial^2E(x', t') \over \partial{x'} \partial{t'}} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'^2}} = 0$$ 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 \(\partial/\partial{x'}\) and \(\partial/\partial{t'}\) by use of the chain rule.) (b) Repeat the analysis of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S'\) the wave equation has the same form as in frame \(S\): $${\partial^2E(x', t') \over \partial{x'}^2} - {1 \over c^2} {\partial^2E(x', t') \over \partial{t'}^2} = 0$$ Explain why this shows that the speed of light in vacuum is c in both frames \(S\) and \(S'\).

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