Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
bartleby

Concept explainers

Question
Book Icon
Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains 10B that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by 10B nuclei and the formation of unstable form of 11B undergoing alpha decay to 7Li is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by 10B and further alpha decay process.

(a)

Expert Solution
Check Mark

Answer to Problem 19.69QP

Solution

The overall equation is given as,

510B+01n37Li+24He

Explanation of Solution

Explanation

The equation representing the absorption of neutron by 10B is given as,

510B+01n511B

The equation for the alpha decay in 11B is given as,

511B37Li+24He

The overall equation is given as,

510B+01n37Li+24He

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron 10 in the obtained reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron 10 is -4.42×10-13J_ .

Explanation of Solution

Explanation

Given

The mass of 10B is 10.0129amu .

The mass of 1n is 1.00866amu .

The mass of 7Li is 7.01600amu .

The mass of 4He is 4.00260amu .

The reaction is,

510B+01n37Li+24He

The formula for the amount of energy released during fusion is given as,

ΔE=Δmc2 (1)

Where,

  • Δm is the change in the mass.
  • c is the speed of the light (3.0×108m/s) .

The change in the mass is calculated as,

Δm=Mass of productsMass of reactants

Substitute the mass of reactants and products in the above equation as,

Δm=Mass of productsMass of reactants=(7.01600amu+4.00260amu)(10.0129amu+1.00866amu)=11.0186amu11.02156amu=0.00296amu

The conversion of amu into kg is done as,

1amu=1.66054×1027kg

Hence, the conversion of 0.00296amu into kg is done as,

0.00296amu=0.00296×1.66054×1027kg=0.0049151984×1027kg

Substitute the value of Δm and c in the equation (1),

ΔE=Δmc2=0.0049151984×1027kg×(3×108m/s)2=4.42×1013kgm2s2

The conversion of kgm2s2 to J is done as,

1kgm2s2=1J

Hence, the conversion of 4.42×1013kgm2s2 to J is done as,

4.42×1013kgm2s2=4.42×1013×1J=-4.42×10-13J_

Therefore, energy released by each nucleus of Boron 10 is -4.42×10-13J_ .

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

(c)

Expert Solution
Check Mark

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

  1. a. The overall equation is given as,

    510B+01n37Li+24He

  2. b. The energy released by each nucleus of Boron 10 is -4.42×10-13J_ .
  3. c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy

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 19 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 19 - Prob. 19.5VPCh. 19 - Prob. 19.6VPCh. 19 - Prob. 19.7VPCh. 19 - Prob. 19.8VPCh. 19 - Prob. 19.9VPCh. 19 - Prob. 19.10VPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - Prob. 19.14QPCh. 19 - Prob. 19.15QPCh. 19 - Prob. 19.16QPCh. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.32QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - Prob. 19.35QPCh. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - Prob. 19.43QPCh. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - Prob. 19.51QPCh. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - Prob. 19.55QPCh. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - Prob. 19.66QPCh. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.85APCh. 19 - Prob. 19.86APCh. 19 - Prob. 19.87APCh. 19 - Prob. 19.88APCh. 19 - Prob. 19.89APCh. 19 - Prob. 19.90APCh. 19 - Prob. 19.91APCh. 19 - Prob. 19.92APCh. 19 - Prob. 19.93APCh. 19 - Prob. 19.94APCh. 19 - Prob. 19.95APCh. 19 - Prob. 19.96APCh. 19 - Prob. 19.97APCh. 19 - Prob. 19.98APCh. 19 - Prob. 19.99APCh. 19 - Prob. 19.100AP
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.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY