/*! 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 13 What nuclide is produced in the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 43: Problem 13

What nuclide is produced in the following radioactive decays? (a) \(\alpha\) decay of \({ }_{94}^{239} \mathrm{Pu} ;\) (b) \(\beta\) decay of \({ }_{11}^{24} \mathrm{Na} ;\) (c) \(\beta^{+}\) decay of \({ }_{8}^{15} \mathrm{O}\).

Short Answer

Expert verified
The nuclides produced in the respective decays are: (a) \({ }_{92}^{235} \mathrm{U}\) (b) \({ }_{12}^{24} \mathrm{Mg}\) (c) \({ }_{7}^{15} \mathrm{N}\).

Step by step solution

01

- Alpha Decay of Pu-239

In an alpha decay, a nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons. Thus for an \(\alpha\) decay of \({ }_{94}^{239} \mathrm{Pu}\), we subtract 2 from the atomic number and 4 from the atomic mass number to obtain the new nuclide. This gives us \({ }_{92}^{235} \mathrm{U}\).
02

- Beta Decay of Na-24

In a beta decay, a neutron is converted into a proton and an electron, which is then emitted. Thus for a \(\beta\) decay of \({ }_{11}^{24} \mathrm{Na}\), we add 1 to the atomic number while the atomic mass number remains unchanged. This gives us \({ }_{12}^{24} \mathrm{Mg}\).
03

- Positron Decay of O-15

In a positron decay, a proton is converted into a neutron and the released positron is then immediately emitted. So for a \(\beta^{+}\) decay of \({ }_{8}^{15} \mathrm{O}\), we subtract 1 from the atomic number and the atomic mass number remains unchanged. This gives us \({ }_{7}^{15} \mathrm{N}\).

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.

Alpha Decay
In alpha decay, a radioactive nucleus emits an alpha particle. Alpha particles are composed of two protons and two neutrons. This means when an alpha particle is emitted, the original atom loses two protons and two neutrons.

Consequently, for each alpha decay:
  • The atomic number of the nuclide decreases by 2.
  • The mass number (sum of protons and neutrons) decreases by 4.
This change transforms the original element into a completely different element within the periodic table. For instance, in the alpha decay of plutonium-239 (\({ }^{239}_{94}\text{Pu}\)), the loss of 2 protons and 4 total nucleons results in uranium-235 (\({ }^{235}_{92}\text{U}\)). This process is a common form of decay for heavy elements and is often accompanied by the emission of energy in the form of radiation.
Beta Decay
During beta decay, a neutron in a nucleus transforms into a proton and emits an electron in the process. This emitted electron is known as a beta particle. As a result:
  • The atomic number increases by 1 since the newly formed proton is added to the atomic structure.
  • The mass number remains unchanged because the total number of nucleons (protons and neutrons) does not change.
For example, when sodium-24 (\({ }^{24}_{11}\text{Na}\)) undergoes beta decay, it transforms into magnesium-24 (\({ }^{24}_{12}\text{Mg}\)) because a neutron has turned into a proton, increasing the atomic number by 1.

Beta decay is a common decay mode for unstable isotopes with an excess of neutrons and results in the formation of new elements.
Positron Emission
Positron emission occurs when a proton in a radioactive nuclide transforms into a neutron, releasing a positron in the process. A positron is the antimatter counterpart of an electron. In this type of decay:
  • The atomic number decreases by 1, reflecting the change from proton to neutron.
  • The mass number remains unchanged, since there is no net loss of nucleons.
An example is oxygen-15 (\({ }^{15}_{8}\text{O}\)), which through positron emission changes into nitrogen-15 (\({ }^{15}_{7}\text{N}\)) as it converts a proton to a neutron, reducing the atomic number by 1.

Positron emission is part of a subclass of beta decay and occurs in isotopes where there is an excess of protons, balancing the internal nuclear forces.

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 radioactive nuclide \({ }^{199} \mathrm{Pt}\) has a half-life of 30.8 minutes. \(\mathrm{A}\) sample is prepared that has an initial activity of \(7.56 \times 10^{11}\) Bq. (a) How many \({ }^{199} \mathrm{Pt}\) nuclei are initially present in the sample? (b) How many are present after 30.8 minutes? What is the activity at this time? (c) Repeat part (b) for a time 92.4 minutes after the sample is first prepared.

The common isotope of uranium, \({ }^{238} \mathrm{U},\) has a half-life of \(4.47 \times 10^{9}\) years, decaying to \({ }^{234} \mathrm{Th}\) by alpha emission. (a) What is the decay constant? (b) What mass of uranium is required for an activity of 1.00 curie? (c) How many alpha particles are emitted per second by \(10.0 \mathrm{~g}\) of uranium?

In an industrial accident a \(65 \mathrm{~kg}\) person receives a lethal whole- body equivalent dose of \(5.4 \mathrm{~Sv}\) from \(\mathrm{x}\) rays. (a) \(\mathrm{What}\) is the equivalent dose in rem? (b) What is the absorbed dose in rad? (c) What is the total energy absorbed by the person's body? How does this amount of energy compare to the amount of energy required to raise the temperature of \(65 \mathrm{~kg}\) of water \(0.010 \mathrm{C}^{\circ} ?\)

Radiation Overdose. If a person's entire body is exposed to \(5.0 \mathrm{~J} / \mathrm{kg}\) of \(\mathrm{x}\) rays, death usually follows within a few days. (a) Express this lethal radiation dose in Gy, rad, Sv, and rem. (b) How much total energy does a \(70.0 \mathrm{~kg}\) person absorb from such a dose? (c) If the \(5.0 \mathrm{~J} / \mathrm{kg}\) came from a beam of protons instead of \(\mathrm{x}\) rays, what would be the answers to parts (a) and (b)?

Tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) is an unstable isotope of hydrogen; its mass, including one electron, is 3.016049 u. (a) Show that tritium must be unstable with respect to beta decay because the decay products ( \({ }_{2}^{3}\) He plus an emitted electron) have less total mass than the tritium. (b) Determine the total kinetic energy (in MeV) of the decay products, taking care to account for the electron masses correctly.
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Quantum Physics

Read Explanation

Classical Mechanics

Read Explanation

Electromagnetism

Read Explanation

Circular Motion and Gravitation

Read Explanation

Electricity

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.