   Chapter 25, Problem 21PS

Chapter
Section
Textbook Problem

Complete the following nuclear equations. Write the mass number and atomic number for the remaining particle, as well as its symbol. (a) 26 54 Fe   +  2 4   He   →   2   0 1 n   +   ?   (b) 13 27 Al   +  2 4   He   →   1S 30 P   +   ?   (c) 16 32 S   +  U 1   n   →   1 1 H   +   ?   (d) 42 96 Mo   +  1 2   H   →   0 1 n   +   ?   (e) 42 98 Mo   +  0 1   n   →   43 99 Tc   +   ?   (f) 9 18 F   →   8 18 O   +   ?

(a)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Concept Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

Explanation

The radioactive isotope of iron-54 when irradiated with alpha particle forms 2856Ni by the emission of 2 neutrons

(b)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(c)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(d)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(e)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

(f)

Interpretation Introduction

Interpretation:

Given reaction has to be completed representing the mass number and atomic number.

Introduction: In this radioactive decay process the unstable isotopes loses their energy by emitting radiation. It is converted to stable isotopes. The emitting radiations are positron emission, gamma emission, beta emission and electron capture.

Mass number is the sum of neutron and protons.

Atomic number is the number of protons.

In alpha decay, there will be lose of He nucleus in which mass number decreases by four and atomic number decreases by two.

In beta decay, there will be a lose of electron from nucleus (neutron turns into proton): there will be no change in mass number and atomic number increases by one.

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 