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To illustrate the principle, let's say we start with a nucleus whose mass is \(200.000\) a.m.u. and inject \(1,600 \mathrm{MeV}\) of energy to completely dismantle the nucleus into its constituent parts. How much mass would the final collection of parts have? a) the exact same: \(200.000\) a.m.u. b) less than \(200.000\) a.m.u. c) more than \(200.000\) a.m.u.

Short Answer

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Question: After injecting a certain amount of energy to dismantle a nucleus, the mass of the final collection of parts is: a) the exact same: 200.000 a.m.u. b) less than 200.000 a.m.u. c) more than 200.000 a.m.u. Please provide the initial mass of the nucleus and the energy injected in MeV.

Step by step solution

01

Understand the mass-energy equivalence formula

The mass-energy equivalence formula, E=mc^2, states that energy (E) is equal to the mass (m) multiplied by the speed of light (c) squared. In this exercise, we are given the energy injected into the nucleus, and we will use this formula to find out the change in mass.
02

Convert the given energy to a.m.u.

The energy given is in \(\mathrm{MeV}\), and we need to convert it to atomic mass units (a.m.u.). First, we need to convert \(\mathrm{MeV}\) to Joules using the conversion factor 1 \(\mathrm{MeV} = 1.602\times10^{-13}\,\mathrm{J}\). Then, we can convert Joules to a.m.u. using the conversion factor 1 a.m.u. = 931.5 \(\mathrm{MeV/c^2}\).
03

Find the change in mass

As we have the energy in a.m.u. now, we can use the mass-energy equivalence formula to find the change in mass: \(\Delta m = \dfrac{E}{c^2}\).
04

Determine the final mass

Add the change in mass to the initial mass (200.000 a.m.u.) to find the final mass of the constituent parts.
05

Compare the final mass to the initial mass

Based on the final mass obtained in Step 4, choose the correct answer among the options a) the exact same: 200.000 a.m.u., b) less than 200.000 a.m.u., or c) more than 200.000 a.m.u.

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Key Concepts

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

Nuclear Physics
Nuclear physics is a fascinating field that explores the fundamental particles and forces at the core of an atom. It specifically looks into the behavior, structure, and interactions of atomic nuclei. At the heart of nuclear physics is the understanding that nuclei are incredibly dense regions, composed mainly of protons and neutrons tightly bound together by strong nuclear forces. When discussing nuclear reactions, whether fission or fusion, these processes involve the rearrangement of these nuclear components.

Understanding nuclear physics is crucial when discussing concepts such as nuclear energy, radioactive decay, and nuclear reactions. Each of these processes relies on the manipulation of the nucleus to release or absorb energy. For instance:
  • In nuclear fission, the nucleus splits into two smaller nuclei, along with the release of energy.
  • In nuclear fusion, two lighter nuclei merge to form a heavier nucleus, releasing energy in the process as well.
These phenomena are directly linked to mass-energy equivalence, highlighting how even small changes in mass can lead to significant energy transformations.
Atomic Mass Unit (a.m.u.)
The atomic mass unit (a.m.u.) is a standard unit of mass that quantifies an atom's mass. It's particularly handy in nuclear physics and chemistry, as it correlates to the mass of protons and neutrons in an atomic nucleus. One atomic mass unit is approximately equal to the mass of one proton or one neutron, which is around \(1.66 \times 10^{-27}\) kilograms.

In practical terms, using a.m.u. simplifies calculations involving atomic-scale masses which are difficult to express in standard units like kilograms. For example, when considering nuclear reactions, using kilograms would lead to unwieldy decimals, while a.m.u. makes these calculations manageable.
  • To convert energy to mass in atomic mass units, there is a key conversion factor: \(1 \text{ a.m.u.} = 931.5 \text{ MeV/c}^2\).
  • This factor allows for a seamless transition between energy (expressed in MeV) and mass (in a.m.u.), enabling us to calculate changes in mass due to energy transformations.
Energy Conversion
Energy conversion is a foundational concept that explains how energy can change forms, be it from kinetic to potential energy, or more relevantly here, between mass and energy. Mass-energy equivalence, given by Einstein's famous equation \(E=mc^2\), establishes that energy (\(E\)) can be directly converted into mass (\(m\)) and vice versa. This is crucial in nuclear physics, where enormous amounts of energy can result from small mass changes.

When discussing nuclear processes, energy conversion often involves:
  • Transforming the released energy into other forms of work or heat.
  • Utilizing the energy absorbed in bridging atomic nuclei or breaking them apart.
For the given exercise, energy conversion becomes essential in understanding how the injected energy of \(1,600 \text{ MeV}\) affects the nucleus’s mass. By converting this energy into a.m.u., we tie back to our first concept, illustrating how these atomic-scale changes manifest in observable differences in nuclear mass. In essence, mass-energy conversions are at the core of how nuclear reactions are both sustained and controlled.

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Most popular questions from this chapter

Operating approximately 450 nuclear plants over about 60 years at a total thermal level of \(1 \mathrm{TW}\), we have had two major radioactive releases into the environment. If we went completely down the nuclear road and get all \(18 \mathrm{TW}^{89}\) this way, what rate of accidents might we expect, if the rate just scales with usage levels?

A large boulder whose mass is \(1,000 \mathrm{~kg}\) having a specific heat capacity of \(1,000 \mathrm{~J} / \mathrm{kg} /{ }^{\circ} \mathrm{C}\) is heated from \(0^{\circ} \mathrm{C}\) to a glowing \(1,800^{\circ} \mathrm{C}\). How much more massive is it, assuming no atoms have been added or subtracted?

Since each nuclear plant delivers \(\sim 1 \mathrm{GW}\) of electrical power, at \(\sim 40 \%\) thermodynamic efficiency this means a thermal generation rate of \(2.5\) GW. How many nuclear plants would we need to supply all 18 TW of our current energy demand? Since a typical lifetime is 50 years before decommissioning, how many days, on average would it be between new plants coming online (while old ones are retired) in a steady state?

A particular fission of \({ }^{235} \mathrm{U}+\mathrm{n}(\) total \(A=236)\) breaks up. One fragment has \(Z=54\) and \(N=86\), making it \({ }^{140} \mathrm{Xe} .\) If no extra neutrons are produced in this event, what must the other fragment be, so all numbers add up? Refer to a periodic table (e.g., Fig. B.1; p. 375 ) to learn which element has the corresponding \(Z\) value, and express the result in the notation \({ }^{\mathrm{A}} \mathrm{X}\).

On balance, considering the benefits and downsides of conventional nuclear fission, where do you come down in terms of support for either terminating, continuing, or expanding our use of this technology? Should we pursue breeder reactors at a large scale? Please justify your conclusion based on the things you consider to be most important.

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