/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 What is the mass excess \(\Delta... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 42: Problem 40

What is the mass excess \(\Delta_{1}\) of \({ }^{1} \mathrm{H}\) (actual mass is \(1.007825 \mathrm{u}\) ) in (a) atomic mass units and (b) \(\mathrm{MeV} / c^{2} ?\) What is the mass excess \(\Delta_{\mathrm{n}}\) of a neutron (actual mass is \(1.008665 \mathrm{u}\) ) in (c) atomic mass units and (d) MeV/c \({ }^{2} ?\) What is the mass excess \(\Delta_{120}\) of \({ }^{120} \mathrm{Sn}\) (actual mass is \(119.902197 \mathrm{u}\) ) in (e) atomic mass units and (f) MeV/c \(^{2} ?\)

Short Answer

Expert verified
Mass excess for \(^{1}\mathrm{H}\) is 0.007825 u (7.29 MeV/c虏); for a neutron, it's 0.008665 u (8.08 MeV/c虏); for \(^{120}\mathrm{Sn}\), it's -0.097803 u (-91.10 MeV/c虏).

Step by step solution

01

Define Mass Excess

The mass excess \( \Delta \) is defined as the difference between the actual mass of the nucleus (given in atomic mass units, u) and its mass number (which is an integer value). This can be expressed as \( \Delta = \text{Actual Mass} - \text{Mass Number} \).
02

Calculate Mass Excess for \(^{1}\mathrm{H}\) (a)

For \(^{1}\mathrm{H}\), the actual mass is \(1.007825\,\mathrm{u}\) and the mass number is 1. The mass excess \( \Delta_1 \) in atomic mass units is calculated as: \[ \Delta_1 = 1.007825 - 1 = 0.007825 \mathrm{u} \]
03

Convert Mass Excess to MeV/c^2 for \(^{1}\mathrm{H}\) (b)

To convert the mass excess from atomic mass units to energy units \( (\mathrm{MeV}/c^2) \), use the conversion factor: 1 atomic mass unit \( (\mathrm{u}) \) is approximately \( 931.5 \mathrm{MeV}/c^2 \). \[ \Delta_1 = 0.007825 \, \mathrm{u} \times 931.5 \frac{\mathrm{MeV}}{c^2} \approx 7.29024 \frac{\mathrm{MeV}}{c^2} \]
04

Calculate Mass Excess for Neutron (c)

For a neutron, the actual mass is \(1.008665\, \mathrm{u}\) and its mass number is 1. The mass excess \( \Delta_{\mathrm{n}} \) in atomic mass units is: \[ \Delta_{\mathrm{n}} = 1.008665 - 1 = 0.008665 \mathrm{u} \]
05

Convert Mass Excess to MeV/c^2 for Neutron (d)

Convert the mass excess \( \Delta_{\mathrm{n}} \) from atomic mass units to \( \mathrm{MeV}/c^2 \) using the conversion factor: \[ \Delta_{\mathrm{n}} = 0.008665 \, \mathrm{u} \times 931.5 \frac{\mathrm{MeV}}{c^2} \approx 8.07888 \frac{\mathrm{MeV}}{c^2} \]
06

Calculate Mass Excess for \(^{120}\mathrm{Sn}\) (e)

For \(^{120}\mathrm{Sn}\), the actual mass is \(119.902197\, \mathrm{u}\) and the mass number is 120. The mass excess \( \Delta_{120} \) in atomic mass units is: \[ \Delta_{120} = 119.902197 - 120 = -0.097803 \mathrm{u} \]
07

Convert Mass Excess to MeV/c^2 for \(^{120}\mathrm{Sn}\) (f)

Convert the mass excess \( \Delta_{120} \) from atomic mass units to \( \mathrm{MeV}/c^2 \) using the conversion factor: \[ \Delta_{120} = -0.097803 \, \mathrm{u} \times 931.5 \frac{\mathrm{MeV}}{c^2} \approx -91.10154 \frac{\mathrm{MeV}}{c^2} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Mass Excess
In nuclear physics, the concept of mass excess plays a crucial role in understanding the stability and energy properties of isotopes. Mass excess is defined as the deviation of the actual atomic mass of a nucleus (measured in atomic mass units, or 'u') from the nearest whole number, which is the mass number of the nucleus. This difference, denoted by \( \Delta \), can be expressed as:
\[ \Delta = \text{Actual Mass} - \text{Mass Number} \]
  • Mass Number: It is the total number of protons and neutrons in an atomic nucleus.
  • Actual Mass: The precise mass measured in atomic mass units.
  • Mass Excess: Indicates how much heavier or lighter an isotopic mass is compared to its mass number.
Atomic Mass Units
Atomic Mass Units (amu or simply 'u') is a standard unit for expressing atomic and molecular weights, providing an intuitive sense of scale when dealing with very small masses of atoms and particles. It is defined precisely as one twelfth of the mass of an unbound neutral carbon-12 atom in its ground state. This unit allows scientists to quantify the very small mass deficiencies or excesses present in nuclear physics problems.
An atomic mass unit is approximately equal to \( 1.66053906660 \times 10^{-27} \) kilograms or the reciprocal of Avogadro鈥檚 number, providing a practical way to express these tiny values encountered in atomic physics.
MeV/c^2
MeV/c^2 is an alternative unit of mass derived from the energy unit MeV (Mega-electron Volt) used in particle physics. In the equation \( E = mc^2 \), mass and energy are interchangeable under certain conditions. Thus, one amu can be converted into energy using the conversion:
\[ 1 \text{ amu} = 931.5 \text{ MeV}/c^2 \]

This conversion gives physicists a convenient way to express mass in terms of energy, which is often more intuitive in the context of nuclear reactions and quantum physics. It simplifies computations involving particle interactions, decay processes, and nuclear reactions, where energy transformation is a fundamental aspect.
Conversion Factors
Conversion factors serve the purpose of translating values from one set of units to another, ensuring compatibility and coherence in scientific measurements. For nuclear physics, one key conversion factor relates atomic mass units and energy units (MeV/c^2).
  • Atomic Mass to Energy Conversion: \( 1 \text{ u} = 931.5 \text{ MeV}/c^2 \)
  • This facilitates calculations involving energy transformations based on mass changes in nuclear reactions.
Using conversion factors correctly ensures precision across scientific disciplines and allows for theoretical concepts to translate into experimental validation, as well as linking energy and mass in equations used in high-energy physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The radionuclide \({ }^{56} \mathrm{Mn}\) has a half-life of \(2.58 \mathrm{~h}\) and is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope \({ }^{55} \mathrm{Mn}\), and the manganese-deuteron reaction that produces \({ }^{56} \mathrm{Mn}\) is $$ { }^{55} \mathrm{Mn}+\mathrm{d} \rightarrow{ }^{56} \mathrm{Mn}+\mathrm{p} $$ If the bombardment lasts much longer than the half-life of \({ }^{56} \mathrm{Mn}\), the activity of the \({ }^{56} \mathrm{Mn}\) produced in the target reaches a final value of \(8.88 \times 10^{10} \mathrm{~Bq}\). (a) At what rate is \({ }^{56} \mathrm{Mn}\) being produced? (b) How many \({ }^{56} \mathrm{Mn}\) nuclei are then in the target? (c) What is their total mass?

Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a neutron and a proton meet (assume both to be almost stationary), they combine and form a deuteron, emitting a gamma ray whose energy is \(2.2233 \mathrm{MeV}\). The masses of the proton and the deuteron are \(1.007276467 \mathrm{u}\) and \(2.013553212 \mathrm{u}\), respectively. Find the mass of the neutron from these data.

When aboveground nuclear tests were conducted, the explosions shot radioactive dust into the upper atmosphere. Global air circulations then spread the dust worldwide before it settled out on ground and water. One such test was conducted in October \(1976 .\) What fraction of the \({ }^{90}\) Sr produced by that explosion still existed in October \(2006 ?\) The half-life of \({ }^{90} \mathrm{Sr}\) is \(29 \mathrm{y}\).

A \(90 \mathrm{~kg}\) person receives a whole-body radiation dose of \(2.4 \times 10^{-4}\) Gy, delivered by alpha particles for which the RBE factor is 12. Calculate (a) the absorbed energy in joules and the dose equivalent in (b) sieverts and (c) rem.

A radiation detector records 9500 counts in \(1.00 \mathrm{~min}\). Assuming that the detector records all decays, what is the activity of the radiation source in (a) becquerels and (b) curies?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Solid State Physics

Read Explanation

Medical Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Classical Mechanics

Read Explanation

Thermodynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.