EBK COLLEGE PHYSICS
EBK COLLEGE PHYSICS
11th Edition
ISBN: 8220103600385
Author: Vuille
Publisher: Cengage Learning US
bartleby

Concept explainers

Question
Book Icon
Chapter 29, Problem 12P

(a)

To determine

The binding energy per nucleon of H12 .

(a)

Expert Solution
Check Mark

Answer to Problem 12P

The binding energy per nucleon of H12 is 1.11MeV/nucleon .

Explanation of Solution

Formula to calculate the binding energy per nucleon is,

BEA=[ZmpNmnm(H12)]c2A

  • mp is the mass of the proton.
  • mn is the mass of the neutron.
  • Z is the atomic number.
  • N is the neutron number.
  • m(H12) is the mass of H12 .
  • c is the speed of light.
  • A is the mass number.

Substitute 1 for Z, 1 for N,1.008625 u for mn , 1.007825 u for mp , 2.014102 u for m(H12) , 931.5 MeV/u for c2 and 2 for A in the above equation to find BE/A .

BEA=[(1)(1.007825u)(1)(1.008625u)(2.014102u)](931.5MeV/u)2=1.11MeV/nucleon

Conclusion:

The binding energy per nucleon of M1224g is 1.11MeV/nucleon .

(b)

To determine

The binding energy per nucleon of H24e .

(b)

Expert Solution
Check Mark

Answer to Problem 12P

The binding energy per nucleon of H24e is 7.07MeV/nucleon .

Explanation of Solution

Formula to calculate the binding energy per nucleon is,

BEA=[ZmpNmnm(H24e)]c2A

  • m(H24e) is the mass of H24e .

Substitute 2 for Z, 2 for N,1.008625 u for mn , 1.007825 u for mp , 4.002603 u for m(H24e) , 931.5 MeV/u for c2 and 4 for A in the above equation to find BE/A .

BEA=[(2)(1.007825u)(2)(1.008625u)(4.002603u)](931.5MeV/u)4=7.07MeV/nucleon

Conclusion:

The binding energy per nucleon of R3785b is 7.07MeV/nucleon .

(c)

To determine

The binding energy per nucleon of F2656e .

(c)

Expert Solution
Check Mark

Answer to Problem 12P

The binding energy per nucleon of F2656e is 8.79MeV/nucleon .

Explanation of Solution

Formula to calculate the binding energy per nucleon is,

BEA=[ZmpNmnm(F2656e)]c2A

  • m(F2656e) is the mass of F2656e .

Substitute 26 for Z, 30 for N,1.008625 u for mn , 1.007825 u for mp , 55.934942 u for m(F2656e) , 931.5 MeV/u for c2 and 56 for A in the above equation to find BE/A .

BEA=[(26)(1.007825u)(30)(1.008625u)(55.934942u)](931.5MeV/u)56=8.79MeV/nucleon

Conclusion:

The binding energy per nucleon of R3785b is 8.79MeV/nucleon .

(d)

To determine

The binding energy per nucleon of U92238 .

(d)

Expert Solution
Check Mark

Answer to Problem 12P

The binding energy per nucleon of U92238 is 7.57MeV/nucleon .

Explanation of Solution

Formula to calculate the binding energy per nucleon is,

BEA=[ZmpNmnm(U92238)]c2A

  • m(U92238) is the mass of U92238 .

Substitute 92 for Z, 146 for N,1.008625 u for mn , 1.007825 u for mp , 238.050783 u for m(U92238) , 931.5 MeV/u for c2 and 238 for A in the above equation.

BEA=[(92)(1.007825u)(146)(1.008625u)(238.050783u)](931.5MeV/u)238=7.57MeV/nucleon

Conclusion:

The binding energy per nucleon of U92238 is 7.57MeV/nucleon

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!
Students have asked these similar questions
Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC
In the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?
Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor  µC 6.00 µF capacitor  µC 3.00 µF capacitor  µC capacitor C  µC

Chapter 29 Solutions

EBK COLLEGE PHYSICS

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.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Text book image
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning