1. Calculate the energy released by the reaction given in sample problem 12 and compare it to the energy released by the alpha decay of californium–252. 25Cf → 1Xe + 19Ru +4 ¿n108Ru + 4 on253Cf → 28Cm + HeCm + ¿He96IsotopeAtomic Mass (u)252Cf98252.081626140Xe139.92165108RU44107.91017248Cm96248.072349He4.001506 Sample Problem 12The spontaneous fission of californium-252 releases xenon-140 andi ruthenium-108. How many neutrons are released in this process?Writing the following reaction, we have252Cf → 140Xe + 10°Ru + ? ¿nAdding the atomic numbers, we could see that the totals are balanced.For the mass numbers252 = 140 + 108 + (x)(1)in which x is the number of neutrons. Solving for x252 = 248 + x =x= 252 – 248 = 4Therefore, there are four neutrons that are products of the reaction.The balanced reaction is252Cf108140y54Xe +Ru + 4 ¿n98

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Answered: 1. Calculate the energy released by the…

1. Calculate the energy released by the reaction given in sample problem 12 and compare it to the energy released by the alpha decay of californium–252.

Living by Chemistry

2nd Edition

ISBN:9781464142314

Author:Angelica M. Stacy

Publisher:Angelica M. Stacy

ChapterU5: Fire: Energy , Thermodynamics, And Oxidation-reduction

SectionU5.116: How Absorbing: Spectroscopy

Problem 2E

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Concept explainers

Writing and Balancing Nuclear Equations

A nuclear equation is a symbolic representation of a nuclear reaction using certain symbols. It is studied in nuclear chemistry and nuclear physics.

Question

1. Calculate the energy released by the reaction given in sample problem 12 and compare it to the energy released by the alpha decay of californium–252.

25Cf → 1Xe + 19Ru +4 ¿n
108Ru + 4 on
253Cf → 28Cm + He
Cm + ¿He
96
Isotope
Atomic Mass (u)
252Cf
98
252.081626
140
Xe
139.92165
108RU
44
107.91017
248Cm
96
248.072349
He
4.001506

Sample Problem 12
The spontaneous fission of californium-252 releases xenon-140 and
i ruthenium-108. How many neutrons are released in this process?
Writing the following reaction, we have
252Cf → 140Xe + 10°Ru + ? ¿n
Adding the atomic numbers, we could see that the totals are balanced.
For the mass numbers
252 = 140 + 108 + (x)(1)
in which x is the number of neutrons. Solving for x
252 = 248 + x =x= 252 – 248 = 4
Therefore, there are four neutrons that are products of the reaction.
The balanced reaction is
252Cf
108
140y
54Xe +
Ru + 4 ¿n
98

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