Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Concept explainers

Question
Book Icon
Chapter 29, Problem 80P

(a)

To determine

The decay scheme for 3890Sr.

(a)

Expert Solution
Check Mark

Answer to Problem 80P

The decay scheme for 3890Sr is 3890Sr3990Y+10e+00ν¯.

Explanation of Solution

The 3890Sr undergoes β decay and emit an electron and changes a neutron in the nucleus to a proton. Thus, increases the charge of nucleus by 1. The mass number of the nucleus is the same all along the process A=90.

From periodic table the element with atomic number Z=39 is calcium. And the symbol for the daughter is 3990Y.

The β decay reaction for 3890Sr is

    3890Sr3990Y+10e+00ν¯

Here, ν¯ is an antineutrino its mass is too low, it is negligible.

Thus, the decay scheme for 3890Sr is 3890Sr3990Y+10e+00ν¯.

(b)

To determine

The activity of 2.0 kg of 3890Sr.

(b)

Expert Solution
Check Mark

Answer to Problem 80P

The activity of 2.0 kg of 3890Sr is 1.0×1016 Bq.

Explanation of Solution

A sample containing 3890Sr with a mass of m=2.0 kg and molar mass of 3890Sr is M=89.9077376 g/mol and its half-life is T1/2=28.8 yr . The Avogadro’s number is NA=6.022×1023 nuclei/mol.

Write the formula for half-life

    T1/2=τln2                                                                                            (I)

Here, T1/2 is the half-life and τ is the time constant.

Write the formula for the number of nuclei

    N=mNAM                                                                                            (II)

Here, N is the number of nuclei, m is the mass, NA is the Avogadro’s number and M is the molar mass.

Write the formula for the activity

    R=Nτ                                                                                                 (III)

Here, R is the activity.

Conclusion:

Substitute equation (I) and (II) in (III) to solve for R0

R=NτR=mNAln2T1/2M

Substitute 2.0 kg for m, 6.022×1023 nuclei/mol for NA, 28.8 yr for T1/2 and 89.9077376 g/mol for M in the above equation to find the value of R. [Note: 1 yr=3.156×107 s]

R=2.0 kg×6.022×1023 nuclei/mol89.9077376 g/mol×28.8 yr×3.156×107 s/yrR=1.0×1016 Bq

Thus, the activity of 2.0 kg of 3890Sr is 1.0×1016 Bq.

(c)

To determine

The activity of 3890Sr after 1000 yr.

(c)

Expert Solution
Check Mark

Answer to Problem 80P

The activity of 3890Sr after 1000 yr is 3.6×105 Bq.

Explanation of Solution

The initial activity of 3890Sr is R0=1.0×1016 Bq, then the after a period of time t=1000 yr the activity R is

Write the formula for the activity

    R=R0et/τ                                                                                            (IV)

Here, R is the activity after time t, R0 is the activity at t=0 and t is the time.

Conclusion:

Substitute equation (I) in (IV) to solve for R

R=R0etln2/T1/2

Substitute 1.0×1016 Bq for R0, 28.8 yr for T1/2 and 1000 yr for t in the above equation to find the value of R

R=1.0×1016 Bq×e1000 yr×ln2/28.8 yrR=3.6×105 Bq

Thus, the activity of 3890Sr after 1000 yr is 3.6×105 Bq.

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!

Chapter 29 Solutions

Physics

Ch. 29.4 - Practice Problem 29.9 The Age of Ötzi In 1991, a...Ch. 29.4 - Prob. 29.10PPCh. 29.5 - Prob. 29.11PPCh. 29.6 - Prob. 29.6CPCh. 29.6 - Prob. 29.12PPCh. 29.7 - Prob. 29.13PPCh. 29.8 - Prob. 29.14PPCh. 29 - Prob. 1CQCh. 29 - Prob. 2CQCh. 29 - Prob. 3CQCh. 29 - Prob. 4CQCh. 29 - Prob. 5CQCh. 29 - Prob. 6CQCh. 29 - Prob. 7CQCh. 29 - Prob. 8CQCh. 29 - Prob. 9CQCh. 29 - Prob. 10CQCh. 29 - Prob. 11CQCh. 29 - Prob. 12CQCh. 29 - Prob. 13CQCh. 29 - Prob. 14CQCh. 29 - Prob. 1MCQCh. 29 - Prob. 2MCQCh. 29 - Prob. 3MCQCh. 29 - Prob. 4MCQCh. 29 - Prob. 5MCQCh. 29 - Prob. 6MCQCh. 29 - Prob. 7MCQCh. 29 - Prob. 8MCQCh. 29 - Prob. 9MCQCh. 29 - Prob. 10MCQCh. 29 - Prob. 1PCh. 29 - Prob. 2PCh. 29 - Prob. 3PCh. 29 - Prob. 4PCh. 29 - Prob. 5PCh. 29 - Prob. 6PCh. 29 - Prob. 7PCh. 29 - Prob. 8PCh. 29 - Prob. 9PCh. 29 - Prob. 10PCh. 29 - Prob. 11PCh. 29 - Prob. 12PCh. 29 - Prob. 13PCh. 29 - Prob. 14PCh. 29 - Prob. 15PCh. 29 - Prob. 16PCh. 29 - Prob. 17PCh. 29 - Prob. 18PCh. 29 - Prob. 19PCh. 29 - Prob. 20PCh. 29 - Prob. 21PCh. 29 - Prob. 22PCh. 29 - Prob. 23PCh. 29 - Prob. 24PCh. 29 - Prob. 25PCh. 29 - Prob. 26PCh. 29 - Prob. 27PCh. 29 - Prob. 28PCh. 29 - Prob. 29PCh. 29 - Prob. 30PCh. 29 - Prob. 31PCh. 29 - Prob. 32PCh. 29 - Prob. 33PCh. 29 - Prob. 34PCh. 29 - Prob. 35PCh. 29 - Prob. 36PCh. 29 - Prob. 37PCh. 29 - Prob. 38PCh. 29 - Prob. 39PCh. 29 - Prob. 40PCh. 29 - Prob. 41PCh. 29 - Prob. 42PCh. 29 - Prob. 43PCh. 29 - Prob. 44PCh. 29 - Prob. 45PCh. 29 - Prob. 46PCh. 29 - Prob. 47PCh. 29 - Prob. 48PCh. 29 - Prob. 49PCh. 29 - Prob. 50PCh. 29 - Prob. 52PCh. 29 - Prob. 51PCh. 29 - Prob. 53PCh. 29 - Prob. 54PCh. 29 - Prob. 55PCh. 29 - Prob. 56PCh. 29 - Prob. 57PCh. 29 - Prob. 58PCh. 29 - Prob. 59PCh. 29 - Prob. 60PCh. 29 - Prob. 61PCh. 29 - Prob. 62PCh. 29 - Prob. 63PCh. 29 - Prob. 64PCh. 29 - Prob. 65PCh. 29 - Prob. 66PCh. 29 - Prob. 67PCh. 29 - Prob. 68PCh. 29 - Prob. 69PCh. 29 - Prob. 70PCh. 29 - Prob. 71PCh. 29 - Prob. 72PCh. 29 - Prob. 73PCh. 29 - Prob. 74PCh. 29 - Prob. 75PCh. 29 - Prob. 76PCh. 29 - Prob. 77PCh. 29 - Prob. 78PCh. 29 - Prob. 79PCh. 29 - Prob. 80PCh. 29 - Prob. 81PCh. 29 - Prob. 82PCh. 29 - Prob. 83PCh. 29 - Prob. 84PCh. 29 - Prob. 85PCh. 29 - Prob. 86PCh. 29 - Prob. 87PCh. 29 - Prob. 88PCh. 29 - Prob. 89PCh. 29 - Prob. 90PCh. 29 - Prob. 91PCh. 29 - Prob. 92PCh. 29 - Prob. 93PCh. 29 - Prob. 94P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON