Concept explainers
(a)
To write: The
(a)
Answer to Problem 4PEB
Solution:
Explanation of Solution
Introduction:
In beta decay process, the atomic number increases by one while there is no effect on mass number..
Explanation:
In order to write the nuclear equation for the beta emission decay of
Step 1: The symbol for beta particle is
Step 2: From the subscripts, atomic number is calculated as:
Step 3:From the superscripts, mass number of nitrogen is calculated as:
Conclusion:
Thus, the complete nuclear equation for the beta emission decay of
(b)
To write: The nuclear equation for the beta emission decay of
(b)
Answer to Problem 4PEB
Solution:
Explanation of Solution
Introduction:
In beta decay process, the atomic number increases by one while there is no effect on mass number.
Explanation:
In order to write the nuclear equation for the beta emission decay of
Step 1: The symbol for beta particle is
Step 2: From the subscripts, atomic number is calculated as:
Step 3: From the superscripts, mass number of nickel is calculated as:
Conclusion:
Thus, the complete nuclear equation for the beta emission decay of
(c)
To write: The nuclear equation for the beta emission decay of
(c)
Answer to Problem 4PEB
Solution:
Explanation of Solution
Introduction:
In beta decay process, the atomic number increases by one while there is no effect on mass number.
Explanation:
In order to write the nuclear equation for the beta emission decay of
Step 1: The symbol for beta particle is
Step 2: From the subscripts, atomic number is calculated as:
Step 3: From the superscripts, mass number of magnesium is calculated as:
Conclusion:
Thus, the complete nuclear equation for the beta emission decay of
(d)
To write: The nuclear equation for the beta emission decay of
(d)
Answer to Problem 4PEB
Solution:
Explanation of Solution
Introduction:
In beta decay process, the atomic number increases by one while there is no effect on mass number.
Explanation:
In order to write the nuclear equation for the beta emission decay of
Step 1: The symbol for beta particle is
Step 2: From the subscripts, atomic number is calculated as:
Step 3: From the superscripts, mass number of americium is calculated as:
Conclusion:
Thus, the complete nuclear equation for the beta emission decay of
(e)
To determine/To Check: The nuclear equation for the beta emission decay of
(e)
Answer to Problem 4PEB
Solution:
Explanation of Solution
Introduction:
In beta decay process, the atomic number increases by one while there is no effect on mass number.
Explanation:
In order to write the nuclear equation for the beta emission decay of
Step 1: The symbol for beta particle is
Step 2: From the subscripts, atomic number is calculated as:
Step 3: From the superscripts, mass number of xenon is calculated as:
Conclusion:
Thus, the complete nuclear equation for the beta emission decay of
(f)
To write: The nuclear equation for the beta emission decay of
(f)
Answer to Problem 4PEB
Solution:
Explanation of Solution
Introduction:
In beta decay process, the atomic number increases by one while there is no effect on mass number.
Explanation:
In order to write the nuclear equation for the beta emission decay of
Step 1: The symbol for beta particle is
Step 2: From the subscripts, atomic number is calculated as:
Step 3: From the superscripts, mass number of bismuth is calculated as:
Conclusion:
Thus, the complete nuclear equation for the beta emission decay of
Want to see more full solutions like this?
Chapter 13 Solutions
Physical Science - With Lab Manual
- (a) Calculate the energy released in the a decay of 238U. (b) What fraction of the mass at a single 238U is destroyed in the decay? The mass of 234Th is 234.043593 u. (c) Although the fractional mass loss is laws for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?arrow_forwardThe ceramic glaze on a red-orange “Fiestaware” plate is U2O3and contains 50.0 grams of 238U, but very little 235U. (a) What is the activity of the plate? (b) Calculate the total energy that will be released by the 238U decay, (c) If energy is worth 12.0 cents per kWh , what is the monetary value of the energy emitted? (These brightly- colored ceramic plates went out of production some 30 years ago, but are still available as collectibles.)arrow_forwardData from the appendices and the periodic table may be needed for these problems. Unreasonable Results A nuclear physicist finds 1.0 (g of 236U in a piece of uranium ore and assumes ii is primordial since its halflife is 2.3107y. (a) Calculate the amount at 236U that would had to have been on Earth when it formed 4.5109y ago for 1.0 (g to be left today. (b) What is unreasonable about this result? (c) What assumption is responsible?arrow_forward
- Data from the appendices and the periodic table may be needed for these problems. The ceramic glaze on a red-orange Fiestaware plate is U2O3 and contains 50.0 grams of 238U, but very little 235U. (a) What is the activity of the plate? (b) Calculate the total energy that will be released by the 238U decay. (c) If energy is worth 12.0 cents per kW (h, what is the monetary value of the energy emitted? (These plates went out of production some 30 years ago, but are still available as collectibles.)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_forwardWhy is the number of neutrons greater than the number of protons in stable nuclei that have an A greater than about 40? Why is this effect more pronounced for the heaviest nuclei?arrow_forward
- (a) Write the decay equation for the decay of 235U. (b) What energy is released in this decay? The mass of the daughter nuclide is 231.036298 u. (c) Assuming the residual nucleus is formed in its ground state, how much energy goes to the particle?arrow_forward(a) Neutron activation of sodium, which is 100% 23Na, produces 24Na, which is used in some heart scans, as seen in Table 32.1. The equation for the reaction is 23Na+n24Na+ . Find its energy output, given the mass of 24Na is 23.990962 u. (b) What mass at 24Na produces the needed 5.0mCi activity, given its halflife is 15.0 h?arrow_forward(a) Calculate the number of grams of deuterium in an 80.000L swimming pool, given deuterium is 0.0150% of natural hydrogen. (b) Find the energy released in joules if this deuterium is fused via the reaction 2H+2H3He+n. (c) Could the neutrons be used to create more energy? (d) Discuss the amount of this type of energy in a swimming pool as compared to that in, say, a gallon of gasoline, also taking into consideration that water is far more abundant.arrow_forward
- College PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegeUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher: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