/*! 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 24 A sample contains one \(\mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 33: Problem 24

A sample contains one \(\mathrm{kg} \mathrm{O}^{19}\) nuclei. The sample decays according to following equation. $$ \mathrm{O}^{19} \longrightarrow \mathrm{F}^{19}+e+\overline{\mathrm{v}} $$ The mass of sample after one half-life period is: (a) lesser than \(1 / 2 \mathrm{~kg}\) (b) equal to \(1 / 2 \mathrm{~kg}\) (c) slightly less than \(1 \mathrm{~kg}\) (d) equal to \(1 \mathrm{~kg}\)

Short Answer

Expert verified
(c) slightly less than 1 kg

Step by step solution

01

Understand the Decay Process

The decay process shown is a beta-plus decay, where an \(\mathrm{O}^{19}\) (oxygen-19) nucleus transforms into an \(\mathrm{F}^{19}\) (fluorine-19) nucleus, emitting a positron (\(e\)) and a neutrino (\(\overline{u}\)). This process conserves mass-energy, meaning the type of atom changes, but the mass remains almost identical.
02

Define Half-life Effects

In radioactive decay, the half-life is the time required for half of the radioactive nuclei in a sample to transform (decay) into another element or isotope. After one half-life, half of the \(\mathrm{O}^{19}\) nuclei will have decayed into \(\mathrm{F}^{19}\), but the overall mass of the sample itself doesn鈥檛 significantly change.
03

Consider Mass Conservation

In nuclear decay, while energy is released, there is no noticeable macroscopic change in mass; the total mass remains nearly the same. This is due to the small mass of emitted particles like electrons and neutrinos compared to nuclei, and the energy-mass equivalence (energy emitted doesn't equate to a macroscopic change in mass here).
04

Answer the Question

Since half of the \(\mathrm{O}^{19}\) changes to \(\mathrm{F}^{19}\) but the total mass remains practically unchanged, the mass should be slightly less than \(1 \mathrm{~kg}\). The slight loss is theoretically due to emitted particles, but negligible in practical scales.

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.

Beta-Plus Decay
In beta-plus decay, a proton in an atomic nucleus is converted into a neutron, resulting in the emission of a positron (which is the antimatter counterpart of an electron) and a neutrino (specifically, an electron neutrino). This transformation occurs because the nucleus seeks a more stable configuration by altering its proton-neutron balance. In the documented equation from the exercise:
  • The Oxygen-19 (\[\mathrm{O}^{19}\]) nucleus transforms into a Fluorine-19 (\[\mathrm{F}^{19}\]) nucleus.
  • Alongside this transformation, a positron (\[e^+\]) and a neutrino (\[\overline{\mathrm{v}}\]) are emitted.

This process does not result in significant changes to the atomic mass of the sample, because the emitted particles have only negligible mass. However, it does change the type of atom present and the atomic structure of the nucleus.
Half-Life
Half-life is a crucial concept in understanding radioactive decay. It is defined as the time required for half of the nuclei in a radioactive sample to decay into their daughter isotopes.
For the \[\mathrm{O}^{19}\] nuclei, a half-life signifies that in one half-life period:
  • Half of the nuclear sample would decay into \[\mathrm{F}^{19}\].
  • The count of the original nuclei is reduced by 50%.
However, the physical mass often remains almost the same, because nuclear transformations involve highly energetic events that don't significantly impact the sample's mass on a macroscopic scale.
Mass-Energy Conservation
The principle of mass-energy conservation is pivotal in nuclear physics. According to this principle, mass and energy are interconnected, described famously by Einstein's formula \[E = mc^2\]. The total energy and mass in a closed system remain constant.
In the context of nuclear decay:
  • While a nuclear reaction may release energy (often in the form of emitted particles and radiation), the total mass-energy sum remains constant. The emissions include small particles such as neutrinos and positrons.
  • The mass reduction noticed when energy is emitted is minuscule compared to the mass of the sample itself.
For the \[\mathrm{O}^{19}\] decay described, the mass is almost unchanged post-decay, due to the conservation laws.
Radioactive Decay
Radioactive decay is a natural, spontaneous process where an unstable atomic nucleus loses energy by emitting radiation. It ultimately transforms into a more stable form, or a different element altogether.
Beta-plus decay is a specific type of this broader phenomenon. Elements undergoing radioactive decay do so to achieve a more stable nuclear configuration. Over time, due to decay:
  • An initial amount of radioactive isotopes decreases, resulting in more stable daughter products.
  • The decay continues until a stable non-radioactive product is formed.
  • Throughout this process, while the number of radioactive atoms decreases, the overall mass tends to remain nearly constant on a perceptible scale.
This means that, in practical terms, samples of radioactive material do not lose mass in any discernible way during decay, which is essential information when answering questions related to changes in mass as posed in the original exercise.

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

In a fusion process, a proton and a neutron combine to give a deuterium nucleus. If \(m_{n}\) and \(m_{p}\) be the masses of neutron and proton respectively, the mass of deuterium nucleus is : (a) equal to \(m_{n}+m_{p}\) (b) more than \(m_{n}+m_{p}\) (c) less than \(m_{n}+m_{p}\) (d) can be less than or more than \(\left(m_{n}+w_{p}\right)\)

Analysis of potassium and argon atoms in a moon rock sample by a mass spectrometer shows that the ratio of the number of stable \(\mathrm{Ar}^{40}\) atoms present to the number of radioactive \(\mathrm{K}^{40}\) atoms is \(7: 1\). Assume that all the argon atoms were produced by the decay of potassium atoms, with a half-life of \(1.25 \times 10^{9}\) year. How old is the rock? (a) \(1.25 \times 10^{9}\) year (b) \(3.75 \times 10^{9}\) year (c) \(8.75 \times 10^{9}\) year (d) \(1.00 \times 10^{10}\) year

For measuring the activity of a radioactive sample, a count rate meter is used. At certain observation, count rate meter recorded 5050 counts per minute but after 10 minutes later, the count rate showed 2300 counts per minute. The disintegration constant \((\lambda)\) is : (a) \(0.065\) per min (b) \(0.078\) per min (c) \(0.24\) per min (d) \(0.868\) per min

Consider : \(X \stackrel{\cdots \alpha}{\longrightarrow} Y \stackrel{-\alpha}{\longrightarrow} Z\) where half- lives of \(X\) and \(Y\) are 2 year and one month respectively. The ratio of atoms of \(X\) and \(Y\), when transient equilibrium \(\left[T_{1 / 2}(X)>T_{1 / 2}(Y)\right]\) has been established, is : (a) \(24: 1\) (b) \(1: 24\) (c) \(23: 1\) (d) \(1: 23\)

In a nuclear fission : (i) in elements of high atomic mass number, energy is released. (ii) linear momentum and total energy are conserved, but not angular momentum. (iii) linear momentum, angular momentum and total energy are conserved. (iv) the probability of neutron being absorbed by a fissionable nucleus, increases when the neutrons are slowed down. (a) (i), (ii) and (iii) are correct (b) (i), (iii) and (iv) are correct (c) (ii), (iii) and (iv) are correct (d) (ii) and (iv) are correct
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Fluids

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Mechanics and Materials

Read Explanation

Measurements

Read Explanation

Astrophysics

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.