Concept explainers
The emitted particle in the decay process of
Answer to Problem 1OQ
Option (c).
Explanation of Solution
In the decay of radioactive nucleus, the daughter nucleus has the same number of nucleon as the parent nucleon, but the atomic number is changed. Such type of radioactive decay is called beta decay. There are two types of beta decay exist, beta minus decay and beta plus decay. A neutron in the nucleus is transformed in to proton and electron in beta minus decay, and a proton is transformed into a neutron and a positron in beta plus decay.
Here the parent nuclei is
Conclusion:
Since emitted particle is an electron, option (c) is correct.
Emitted particle is an electron. Thus, option (a) is incorrect.
Emitted particle is an electron. Thus, option (b) is incorrect.
Emitted particle is an electron. Thus, option (d) is incorrect.
Emitted particle is an electron. Thus, option (e) is incorrect.
Want to see more full solutions like this?
Chapter 30 Solutions
Principles of Physics
- Suppose 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_forwardundergoes alpha decay, (a) Write the reaction equation, (b) Find the energy released in the decay.arrow_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
- A rare decay mode has been observed in which 222Raemits a 14C nucleus. (a) The decay equation is 222RaAX+14C . Identify the nuclide AX. (b) Find the energy emitted in the decay. The mass of 222Ra is 222.015353 u.arrow_forward(a) A cancer patient is exposed to rays from a 5000Ci 60Co transillumination unit for 32.0 s. The rays are collimated in such a manner that only 1.00% of them strike the patient. Of those, 20.0% are absorbed in a tumor having a mass of 1.50 kg. What is the dose in rem to the tumor, it the average energy per decay is 1.25 MeV? None of the s from the decay reach the patient. (b) Is the dose consistent with stated therapeutic doses?arrow_forwardDerive an approximate relationship between the energy of (decay and halflife using the following data. It may be useful to graph the leg t1/2 against Ea to find some straightline relationship. Table 31.3 Energy and HalfLife for (Decay Nuclide E( (MeV) t1/2 216Ra 9.5 0.18 (s 194Po 7.0 0.7 s 240Cm 6.4 27 d 226Ra 4.91 1600 y 232Th 4.1 1.41010yarrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
- College PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegeModern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning