Skip to main content

Science Chemistry SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED

The following nuclear equations describing the preparation of isotopes used for nuclear medicine: S 32 + n 1 → ? + H 1 M n 55 + H 1 → F e 52 + ? A s 75 + ? → B r 77 X e 124 + n 1 → ? → I 125 + ? Solution: S 16 32 + n 0 1 → P 15 32 + H 1 1 M n 25 55 + H 1 → F e 26 52 + 4 n 0 1 A s 33 75 + 2 H 1 1 → B r 35 77 X e 124 + n 1 → ? → I 125 + + β - 1 0 Explanation: For the balanced nuclear reaction, the sum of charges, i.e., atomic numbers (subscript) and mass numbers (superscript) must be equal on both sides. If the atomic number and mass number of the rest of the particles is known, then we can determine the missing particle in the balanced nuclear reaction. S 32 + n 1 → ? + H 1 The atomic number of S (sulfur) is 16 . In this nuclear equation, the sum of mass numbers on the left side is 33 and on right side is 1 . The sum of atomic numbers (charge) on the left side is 16 and on the right side is 1 . So, to balance the equation, we need the nuclide (or subatomic particle) with the mass number 32 and atomic number 15 . The element with atomic number 15 is Phosphorus ( P ) . The equation will be S 32 + n 1 → P 32 + H 1 Now the mass number is not balanced, so add the coefficient 2 in front of n 0 1 . Therefore, the balanced equation is S 16 32 + n 0 1 → P 15 32 + H 1 1 M n 55 + H 1 → F e 52 + ? The atomic number of M n is 25 ; that of H is 1 , and that of F e is 26 . The sum of mass numbers on left is 56 and on the right side is 52 , so the missing particle on the right side must have the mass number 4 but atomic number 0 . The sum of atomic numbers or charges is 26 on both sides, so the particle must have charge 0 (i.e., no charge). The subatomic particle with the mass number 1 and charge 0 is n 0 1 . Hence the balanced nuclear equation is M n 25 55 + H 1 → F e 26 52 + 4 n 0 1 A s 75 + ? → B r 77 The atomic number of A s is 33 , and that of B r is 35 . The sum of mass number on the left is 77 , and that on the right side is 77 . So the missing particle on the left side must have the mass number 2 . The sum of atomic numbers (charge) on the left side is 33 and on the right side is 35 . So, to balance the charge, we require the missing nuclide with atomic number 2 . The missing particle cannot be Helium because it has mass number 4 . Therefore, we will need two particles of proton of hydrogen. The balanced equation is A s 33 75 + 2 H 1 1 → B r 35 77 X e 54 124 + n 0 1 → ? → I 53 125 + ? The atomic number of Xe is 54 , and that of I is 53 . The sum of mass numbers on the left side is 125 and on the right side is 125 . So the mass numbers are balanced. The sum of atomic numbers (charges) on the left side is 54 since the charge on n (neutron) is 0 and on the right side is 53 , so the missing particle must have the atomic number or charge equal to - 1 . The subatomic particle with mass number 0 and charge - 1 is proton, represented as β - 1 0 . First it forms X e 54 125 and then it emits beta particle to form iodine-125. Therefore, the balanced nuclear reaction is X e 54 124 + n 0 1 → ? → I 53 125 + + β - 1 0 Conclusion: We determined the sum of atomic numbers and mass numbers on both sides of the reaction to identify the missing particle or nuclide that balanced the nuclear equations.

The following nuclear equations describing the preparation of isotopes used for nuclear medicine: S 32 + n 1 → ? + H 1 M n 55 + H 1 → F e 52 + ? A s 75 + ? → B r 77 X e 124 + n 1 → ? → I 125 + ? Solution: S 16 32 + n 0 1 → P 15 32 + H 1 1 M n 25 55 + H 1 → F e 26 52 + 4 n 0 1 A s 33 75 + 2 H 1 1 → B r 35 77 X e 124 + n 1 → ? → I 125 + + β - 1 0 Explanation: For the balanced nuclear reaction, the sum of charges, i.e., atomic numbers (subscript) and mass numbers (superscript) must be equal on both sides. If the atomic number and mass number of the rest of the particles is known, then we can determine the missing particle in the balanced nuclear reaction. S 32 + n 1 → ? + H 1 The atomic number of S (sulfur) is 16 . In this nuclear equation, the sum of mass numbers on the left side is 33 and on right side is 1 . The sum of atomic numbers (charge) on the left side is 16 and on the right side is 1 . So, to balance the equation, we need the nuclide (or subatomic particle) with the mass number 32 and atomic number 15 . The element with atomic number 15 is Phosphorus ( P ) . The equation will be S 32 + n 1 → P 32 + H 1 Now the mass number is not balanced, so add the coefficient 2 in front of n 0 1 . Therefore, the balanced equation is S 16 32 + n 0 1 → P 15 32 + H 1 1 M n 55 + H 1 → F e 52 + ? The atomic number of M n is 25 ; that of H is 1 , and that of F e is 26 . The sum of mass numbers on left is 56 and on the right side is 52 , so the missing particle on the right side must have the mass number 4 but atomic number 0 . The sum of atomic numbers or charges is 26 on both sides, so the particle must have charge 0 (i.e., no charge). The subatomic particle with the mass number 1 and charge 0 is n 0 1 . Hence the balanced nuclear equation is M n 25 55 + H 1 → F e 26 52 + 4 n 0 1 A s 75 + ? → B r 77 The atomic number of A s is 33 , and that of B r is 35 . The sum of mass number on the left is 77 , and that on the right side is 77 . So the missing particle on the left side must have the mass number 2 . The sum of atomic numbers (charge) on the left side is 33 and on the right side is 35 . So, to balance the charge, we require the missing nuclide with atomic number 2 . The missing particle cannot be Helium because it has mass number 4 . Therefore, we will need two particles of proton of hydrogen. The balanced equation is A s 33 75 + 2 H 1 1 → B r 35 77 X e 54 124 + n 0 1 → ? → I 53 125 + ? The atomic number of Xe is 54 , and that of I is 53 . The sum of mass numbers on the left side is 125 and on the right side is 125 . So the mass numbers are balanced. The sum of atomic numbers (charges) on the left side is 54 since the charge on n (neutron) is 0 and on the right side is 53 , so the missing particle must have the atomic number or charge equal to - 1 . The subatomic particle with mass number 0 and charge - 1 is proton, represented as β - 1 0 . First it forms X e 54 125 and then it emits beta particle to form iodine-125. Therefore, the balanced nuclear reaction is X e 54 124 + n 0 1 → ? → I 53 125 + + β - 1 0 Conclusion: We determined the sum of atomic numbers and mass numbers on both sides of the reaction to identify the missing particle or nuclide that balanced the nuclear equations.

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED

2nd Edition

ISBN: 9780393644777

Author: Gilbert

Publisher: MCGRAW-HILL HIGHER EDUCATION

See similar textbooks

Concept explainers

Writing and Balancing Nuclear Equations

A nuclear equation is a symbolic representation of a nuclear reaction using certain symbols. It is studied in nuclear chemistry and nuclear physics.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

Blurred answer

Chapter 21, Problem 21.52QA

Chapter 21, Problem 21.54QA

Chapter 21 Solutions

SMARTWORKS FOR CHEMISTRY: ATOMS FOCUSED

Ch. 21 - Prob. 21.1VP Ch. 21 - Prob. 21.2VP Ch. 21 - Prob. 21.3VP Ch. 21 - Prob. 21.4VP Ch. 21 - Prob. 21.5VP Ch. 21 - Prob. 21.6VP Ch. 21 - Prob. 21.7VP Ch. 21 - Prob. 21.8VP Ch. 21 - Prob. 21.9VP Ch. 21 - Prob. 21.10VP

Ch. 21 - Prob. 21.11VP Ch. 21 - Prob. 21.12VP Ch. 21 - Prob. 21.13QA Ch. 21 - Prob. 21.14QA Ch. 21 - Prob. 21.15QA Ch. 21 - Prob. 21.16QA Ch. 21 - Prob. 21.17QA Ch. 21 - Prob. 21.18QA Ch. 21 - Prob. 21.19QA Ch. 21 - Prob. 21.20QA Ch. 21 - Prob. 21.21QA Ch. 21 - Prob. 21.22QA Ch. 21 - Prob. 21.23QA Ch. 21 - Prob. 21.24QA Ch. 21 - Prob. 21.25QA Ch. 21 - Prob. 21.26QA Ch. 21 - Prob. 21.27QA Ch. 21 - Prob. 21.28QA Ch. 21 - Prob. 21.29QA Ch. 21 - Prob. 21.30QA Ch. 21 - Prob. 21.31QA Ch. 21 - Prob. 21.32QA Ch. 21 - Prob. 21.33QA Ch. 21 - Prob. 21.34QA Ch. 21 - Prob. 21.35QA Ch. 21 - Prob. 21.36QA Ch. 21 - Prob. 21.37QA Ch. 21 - Prob. 21.38QA Ch. 21 - Prob. 21.39QA Ch. 21 - Prob. 21.40QA Ch. 21 - Prob. 21.41QA Ch. 21 - Prob. 21.42QA Ch. 21 - Prob. 21.43QA Ch. 21 - Prob. 21.44QA Ch. 21 - Prob. 21.45QA Ch. 21 - Prob. 21.46QA Ch. 21 - Prob. 21.47QA Ch. 21 - Prob. 21.48QA Ch. 21 - Prob. 21.49QA Ch. 21 - Prob. 21.50QA Ch. 21 - Prob. 21.51QA Ch. 21 - Prob. 21.52QA Ch. 21 - Prob. 21.53QA Ch. 21 - Prob. 21.54QA Ch. 21 - Prob. 21.55QA Ch. 21 - Prob. 21.56QA Ch. 21 - Prob. 21.57QA Ch. 21 - Prob. 21.58QA Ch. 21 - Prob. 21.59QA Ch. 21 - Prob. 21.60QA Ch. 21 - Prob. 21.61QA Ch. 21 - Prob. 21.62QA Ch. 21 - Prob. 21.63QA Ch. 21 - Prob. 21.64QA Ch. 21 - Prob. 21.65QA Ch. 21 - Prob. 21.66QA Ch. 21 - Prob. 21.67QA Ch. 21 - Prob. 21.68QA Ch. 21 - Prob. 21.69QA Ch. 21 - Prob. 21.70QA Ch. 21 - Prob. 21.71QA Ch. 21 - Prob. 21.72QA Ch. 21 - Prob. 21.73QA Ch. 21 - Prob. 21.74QA Ch. 21 - Prob. 21.75QA Ch. 21 - Prob. 21.76QA Ch. 21 - Prob. 21.77QA Ch. 21 - Prob. 21.78QA Ch. 21 - Prob. 21.79QA Ch. 21 - Prob. 21.80QA Ch. 21 - Prob. 21.81QA Ch. 21 - Prob. 21.82QA Ch. 21 - Prob. 21.83QA Ch. 21 - Prob. 21.84QA Ch. 21 - Prob. 21.85QA Ch. 21 - Prob. 21.86QA Ch. 21 - Prob. 21.87QA Ch. 21 - Prob. 21.88QA Ch. 21 - Prob. 21.89QA Ch. 21 - Prob. 21.90QA Ch. 21 - Prob. 21.91QA Ch. 21 - Prob. 21.92QA Ch. 21 - Prob. 21.93QA Ch. 21 - Prob. 21.94QA Ch. 21 - Prob. 21.95QA Ch. 21 - Prob. 21.96QA Ch. 21 - Prob. 21.97QA Ch. 21 - Prob. 21.98QA Ch. 21 - Prob. 21.99QA Ch. 21 - Prob. 21.100QA Ch. 21 - Prob. 21.101QA Ch. 21 - Prob. 21.102QA Ch. 21 - Prob. 21.103QA Ch. 21 - Prob. 21.104QA Ch. 21 - Prob. 21.105QA Ch. 21 - Prob. 21.106QA Ch. 21 - Prob. 21.107QA Ch. 21 - Prob. 21.108QA Ch. 21 - Prob. 21.109QA Ch. 21 - Prob. 21.110QA Ch. 21 - Prob. 21.111QA Ch. 21 - Prob. 21.112QA Ch. 21 - Prob. 21.113QA Ch. 21 - Prob. 21.114QA Ch. 21 - Prob. 21.115QA Ch. 21 - Prob. 21.116QA Ch. 21 - Prob. 21.117QA Ch. 21 - Prob. 21.118QA Ch. 21 - Prob. 21.119QA Ch. 21 - Prob. 21.120QA Ch. 21 - Prob. 21.121QA Ch. 21 - Prob. 21.122QA Ch. 21 - Prob. 21.123QA Ch. 21 - Prob. 21.124QA Ch. 21 - Prob. 21.125QA Ch. 21 - Prob. 21.126QA Ch. 21 - Prob. 21.127QA Ch. 21 - Prob. 21.128QA Ch. 21 - Prob. 21.129QA Ch. 21 - Prob. 21.130QA

Knowledge Booster

Background pattern image

Chemistry

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this 239Pu was the capture of neutrons by 238U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7109 years ago?

Recommended textbooks for you

Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Living By Chemistry: First Edition Textbook
Chemistry
ISBN:9781559539418
Author:Angelica Stacy
Publisher:MAC HIGHER
Introductory Chemistry: An Active Learning Approa...
Chemistry
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Cengage Learning
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co

Text book image

Introductory Chemistry: A Foundation

Chemistry

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Cengage Learning

Text book image

Chemistry: The Molecular Science

Chemistry

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:Cengage Learning

Text book image

Living By Chemistry: First Edition Textbook

Chemistry

ISBN:9781559539418

Author:Angelica Stacy

Publisher:MAC HIGHER

Text book image

Introductory Chemistry: An Active Learning Approa...

Chemistry

ISBN:9781305079250

Author:Mark S. Cracolice, Ed Peters

Publisher:Cengage Learning

Text book image

Chemistry by OpenStax (2015-05-04)

Chemistry

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:OpenStax

Text book image

Chemistry: Matter and Change

Chemistry

ISBN:9780078746376

Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom

Publisher:Glencoe/McGraw-Hill School Pub Co

SEE MORE TEXTBOOKS