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Chapter 37: Problem 58

A photon with energy \(E\) is emitted by an atom with mass \(m\) which recoils in the opposite direction. (a) Assuming that the motion of the atom can be treated nonrelativistically, compute the recoil speed of the atom. (b) From the result of part (a), show that the recoil speed is much less than \(c\) whenever \(E\) is much less than the rest energy \(m c^{2}\) of the atom.

Short Answer

Expert verified
The recoil speed of the atom is \(v = \frac{E}{mc}\), which is much less than \(c\) if \(E \ll mc^2\).

Step by step solution

01

Understand the Energy Conservation Scenario

When the photon is emitted, the energy of the emitted photon is denoted by \(E\). By conservation of energy and momentum, the atom gains kinetic energy which affects its speed. Our aim is to find the recoil speed \(v\) of the atom.
02

Apply Momentum Conservation

Conservation of momentum dictates that the momentum before and after emission must be equal. Initially, the total momentum is zero. After emission, the momentum of the photon is \(\frac{E}{c}\) (as photons have momentum \(p = \frac{E}{c}\)), and the atom's momentum \(mv\) in the opposite direction ensures momentum conservation: \[mv = \frac{E}{c}\].
03

Solve for Recoil Speed

From the momentum conservation equation \(mv = \frac{E}{c}\), solve for the recoil speed \(v\): \[v = \frac{E}{mc}\].
04

Analyze the Recoil Speed Compared to Speed of Light

For part (b), we need to show that the recoil speed is much less than \(c\). If \(E \ll mc^2\), then \(\frac{E}{mc} \ll c\), implying \(v \ll c\), as required.

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Key Concepts

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

Momentum Conservation
Momentum conservation is a fundamental principle in physics stating that the total momentum of a closed system remains constant, provided there are no external forces. In the context of our problem, when a photon is emitted by the atom, the system's total initial momentum is zero. Before the emission, both the atom and photon are considered part of a single system. The photon, upon being emitted, carries away energy and momentum - The photon's momentum is given by \( \frac{E}{c} \), where \( E \) is the energy of the photon and \( c \) is the speed of light. - To keep the system's momentum balanced, the atom must recoil with an equal and opposite momentum.Thus, the atom gains momentum in the opposite direction, defined as \( mv \), where \( m \) is the atom's mass and \( v \) is the recoil speed. The equality in momentum: \[ mv = \frac{E}{c} \] is essential for understanding how energy and momentum distribute between the photon and the atom after emission.
Recoil Speed
Recoil speed is the speed at which an object moves in response to force exerted on it during an interaction. In our scenario, the interaction is the emission of a photon by the atom. The atom gains recoil speed due to the conservation of momentum, which is dictated by the equality \( mv = \frac{E}{c} \).To solve for the recoil speed \( v \), one can rearrange the momentum conservation equation: \[ v = \frac{E}{mc} \] This equation shows us how the photon emission influences the atom's motion. The result indicates the recoil speed is directly proportional to the photon's energy and inversely proportional to the atom's mass and the speed of light. Therefore, the larger the photon's energy or the lighter the atom, the greater the recoil speed. Importantly, this relationship allows us to analyze how significantly an atom's movement is affected by the process of emitting a photon.
Nonrelativistic Motion
Nonrelativistic motion refers to a regime of movement where speeds are much lower than the speed of light, such that relativistic effects (those described by Einstein's theory of relativity) can be ignored. In our case, the problem explicitly asks us to consider nonrelativistic conditions for calculating the recoil speed.Given the equation \( v = \frac{E}{mc} \), to ensure the motion remains nonrelativistic, it is required that \( v \ll c \), meaning the recoil speed is a fraction much less than the speed of light. The condition for this can be assured by showing that the photon's energy \( E \) is much less than the rest energy of the atom, \( mc^2 \).- When \( E \ll mc^2 \), it implies \( \frac{E}{mc} \ll c \), thus ensuring that the recoil speed \( v \) stays within nonrelativistic limits.This consideration is crucial for verifying that classic mechanics can be used without accounting for relativistic effects, simplifying the calculations and understanding.

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Most popular questions from this chapter

A \(60.0-\mathrm{kg}\) person is standing at rest on level ground. How fast would she have to run to (a) double her total energy and (b) increase her total energy by a factor of 10\(?\)

A \(0.100-\mu g\) speck of dust is accelerated from rest to a speed of 0.900\(c\) by a constant \(1.00 \times 10^{6} \mathrm{N}\) force. (a) If the nonrelativistic form of Newton's second law \((\Sigma F=m a)\) is used, how far does-the object travel to reach its final speed? (b) Using the correct relativistic treatment of Section 37.8 , how far does the object travel to reach its final speed? (c) Which distance is greater? Why?

(a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-v t\) 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 $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{2}}=0$$ This has a different form than the wave equation in \(S\) . Hence the Galiean 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 part (a), but use the Lorentz coordinate transformations, Eqs. (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 \(S^{\prime} .\)

A spaceship moving at constant speed \(u\) relative to us broadcasts a radio signal at constant frequency \(f_{0}\) . As the spaceship approaches us, we receive a higher frequency \(f\) ; after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0}\) and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?(\text {Hint} \text { . In this case, successive wave crests }\) move the same distance to the observer and so they have the same transit time. Thus \(f\) equals \(1 / T .\) Use the time dilation formula to relate the periods in the stationary and moving frames) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758 c relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?(\mathrm{c})\) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is \(\lambda=656.3 \mathrm{nm},\) in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler-shifted to \(\lambda=953.4 \mathrm{nm}\) , in the infrared portion of the spectrum. How fast are the emitting atoms moving relative to the earth? Are they approaching the earth or receding from it?
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