Concept explainers
(a)
The number of alpha particles in decay chain starting from U (238) to Pb (206) and U (235) to Pb (207).
Explanation of Solution
Given:
The decay chain of U (238) and U (235) is given as,
Calculation:
In the decay chain starting from U (238) to Pb (206).
As we know the mass of alpha particle is 4.
Let there are x number of alpha particles in a decay chain.
Thus, there are 8 alpha particles.
In the decay chain starting from U (235) to Pb (207).
As we know the mass of alpha particle is 4.
Let there are x number of alpha particles in a decay chain.
Thus, there are 7 alpha particles.
Conclusion:
In decay chain U (238) to Pb (206) there are 8 alpha particles
In decay chain U (235) to Pb (207) there are 7 alpha particles.
(b)
The ratio of number of Uranium-238 to lead-206.
Explanation of Solution
Given:
Formula used:
Number of radioactive nuclei at time t is given by,
Where,
t = time of sample
Calculation:
For U (238), number of Uranium-238 at time t is given by,
For Pb (206), number of lead-206 at time t is given by
Conclusion:
The ratio of number of Uranium-238 to lead-206 is,
(c)
The ratio of number of Uranium-235 to lead-205.
Explanation of Solution
Given:
Formula used:
Number of radioactive nuclei at time t is given by,
Where,
t = time of sample
Calculation:
For U (238), Number of radioactive nuclei at time t is given by,
Conclusion:
The ratio of number of Uranium-235 to lead-207 is
(d)
The ratio of number of lead-206 to lead-207.
Explanation of Solution
Formula used:
Number of radioactive nuclei at time t is given by,
Where,
t = time of sample
Calculation:
Lead-206 and lead-207 are the most stable atoms. Generally, these atoms do not decay at all.
Thus, the ratio of number of Pb (206) nuclei and number of Pb (207) will be equal to the number of atoms of Pb (206) nuclei and number of Pb (207) present initially.
Conclusion:
The ratio of number of lead-206 to lead-207 is
Want to see more full solutions like this?
Chapter 11 Solutions
Inquiry into Physics
- Enter the correct nuclide symbol in each open tan rectangle in Figure P43.25, which shows the sequences of decays in the natural radioactive series starting with the long-lived isotope uranium-235 and ending with the stable nucleus lead-207. Figure P43.25arrow_forwardSuppose you have a pure radioactive material with a half-life of T1/2. You begin with N0 undecayed nuclei of the material at t = 0. At t=12T1/2, how many of the nuclei have decayed? (a) 14N0 (b) 12N0(C) 34N0 (d) 0.707N0 (e) 0.293N0arrow_forward(a) Calculate the energy released in the neutron- Induced fission reaction n+235U92Kr+142Ba+2n , given m(92Kr) = 91.926269 u and m(142Ba)= 141.916361 u. (b) Confirm that the total number of nucleons and total charge are conserved in this reaction.arrow_forward
- Which of the following is the correct daughter nucleus associated with the alpha decay of 72157Hf? (a) 72153Hf (b) 70153Yb (c) 70157Ybarrow_forward(a) Calculate the energy released in the a decay of 238U . (b) What fraction of the mass of a single 238U is destroyed in the decay? The mass of 234Th is 234.043593 u. (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?arrow_forward
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStaxGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill