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Chapter 39: Problem 75

A \(^{57}\) Fe nucleus at rest emits a 14.0-keV photon. Use conservation of energy and momentum to deduce the kinetic energy of the recoiling nucleus in electron volts. (Use \(M c^{2}=8.60 \times 10^{-9} \mathrm{J}\) for the final state of the \(^{57} \mathrm{Fe}\) nucleus.)

Short Answer

Expert verified
The kinetic energy of the recoiling \(^{57}Fe\) nucleus is approximately 91.1 eV.

Step by step solution

01

Translate photon's energy to Joules

The energy of the emitted photon is given as 14.0 keV. To convert it into Joules, multiply this number by \(1.602 \times 10^{-19} \mathrm{J} / \mathrm{eV}\), the conversion ratio: \(E_{\gamma} = 14.0 \times 10^{3} \mathrm{eV} \times 1.602 \times 10^{-19} \mathrm{J} / \mathrm{eV} = 2.24 \times 10^{-15} \mathrm{J}\).
02

Apply conservation of momentum

For the system of nucleus and photon, initial momentum is zero because the nucleus was at rest before emission. Afterwards, momenta of the nucleus and the photon must be equal in magnitude and opposite in direction for the total momentum to remain zero. Since photon鈥檚 momentum equals its energy divided by speed of light \(c\), we get: \(p_{nucleus} = p_{\gamma} = E_{\gamma} / c\).
03

Calculate nucleus' kinetic energy

The kinetic energy of the nucleus can be calculated using the momentum-energy relationship for a non-relativistic particle, \(KE_n = p_{nucleus}^2 / (2M)\). Inserting \(p_{nucleus} = E_{\gamma} / c\), we get: \(KE_n = (E_{\gamma} / c)^2 / (2M)\).
04

Substitute values and solve

Substituting the calculated \(E_{\gamma}\) and the given \(M c^{2}=8.60 \times 10^{-9} \mathrm{J}\), we get: \(KE_n = (2.24 \times 10^{-15})^2 / 2\times c \times (8.60 \times 10^{-9} / c^2)\). Solving results in \(KE_n = 1.46 \times 10^{-23} \mathrm{J}\).
05

Convert to electron volts

Finally, to express the kinetic energy in electron volts (eV), we divide the result in Joules by \(1.602 \times 10^{-19} \mathrm{J} / \mathrm{eV}\), resulting in \( KE_n = 1.46 \times 10^{-23} \mathrm{J} / 1.602 \times 10^{-19} \mathrm{J} / \mathrm{eV} = 91.1 \mathrm{eV}\).

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Key Concepts

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

Conservation of Energy
When we talk about the conservation of energy in nuclear physics, we're referencing one of physics' most fundamental and universal principles: energy cannot be created or destroyed, only transformed from one form to another. This means the total energy of an isolated system remains constant over time.

In our exercise, the system comprised of the ucleus of an iron atom and the photon it emits. The energy before the emission (when the nucleus is at rest) must equal the total energy after the emission 鈥� the energy of the photon plus the kinetic energy of the recoiling nucleus.

This approach simplifies problem solving significantly. By equating the initial and final energy states, we're able to infer unknown values, in this case, the kinetic energy of the nucleus post-photon emission. It's equally important to convert energy values to a common unit for accurate calculations, such as from electron volts to Joules, to maintain consistency throughout the problem.
Conservation of Momentum
Momentum, a fundamental conserved quantity in physics, is conserved in the absence of external forces. For our nuclear physics problem, this means that the initial momentum of the system (which is zero since the nucleus was originally at rest) is the same as the final momentum of the system after the photon has been emitted.

To find the recoiling nucleus's momentum, we acknowledge that it must be equal in magnitude and opposite in direction to the photon's momentum to maintain the system's initial state of zero total momentum. The photon's momentum is calculated by dividing its energy by the speed of light.

Understanding conservation of momentum is crucial for solving nuclear physics problems as it allows us to determine the velocities or kinetic energy of particles post-interaction without any external input of force or energy.
Kinetic Energy Calculation
Kinetic energy is the energy an object possesses due to its motion. In the context of our nuclear physics problem, we calculate the kinetic energy of the recoiling nucleus using its momentum, determined under the conservation of momentum concept. The kinetic energy (KE) is given by the equation (KE = p^2 / (2M)), where 'p' is momentum and 'M' is the mass of the nucleus.

After we equate the nucleus's momentum to the photon's momentum, the kinetic energy calculation comes down to plugging the quantities we've determined or were given into this equation. The result is a precise measure of the nucleus's kinetic energy in Joules, which we then convert back to electron volts for coherence with the original problem statement.

Such calculations underscore the interconnectedness of physical principles鈥攈ow a change in momentum reflects as kinetic energy鈥攃larifying the motion and energetic transitions of particles in nuclear reactions.

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Most popular questions from this chapter

According to observer \(\mathrm{A}\), two objects of equal mass and moving along the \(x\) axis collide head on and stick to each other. Before the collision, this observer measures that object 1 moves to the right with a speed of \(3 c / 4\) while object 2 moves to the left with the same speed. According to observer \(\mathrm{B}\), however, object 1 is initially at rest. (a) Determine the speed of object 2 as seen by observer B. (b) Compare the total initial energy of the system in the two frames of reference.

A Klingon spacecraft moves away from the Earth at a speed of \(0.800 c\) (Fig. \(\mathrm{P} 39.26\) ). The starship Enterprise pursues at a speed of \(0.900 c\) relative to the Earth. Observers on the Earth see the Enterprise overtaking the Klingon craft at a relative speed of \(0.100 c .\) With what speed is the Enterprise overtaking the Klingon craft as seen by the crew of the Enterprise?

An unstable particle at rest breaks into two fragments of unequal mass. The mass of the first fragment is \(2.50 \times 10^{-28} \mathrm{kg},\) and that of the other is \(1.67 \times 10^{-27} \mathrm{kg}.\) If the lighter fragment has a speed of \(0.893 c\) after the breakup, what is the speed of the heavier fragment?

In a laboratory frame of reference, an observer notes that Newton's second law is valid. Show that it is also valid for an observer moving at a constant speed, small compared with the speed of light, relative to the laboratory frame.

The identical twins Speedo and Goslo join a migration from the Earth to Planet X. It is 20.0 ly away in a reference frame in which both planets are at rest. The twins, of the same age, depart at the same time on different spacecraft. Speedo's craft travels steadily at \(0.950 c,\) and Goslo's at \(0.750 c .\) Calculate the age difference between the twins after Goslo's spacecraft lands on Planet X. Which twin is the older?
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