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.

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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
Transcribed Image Text: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
Transcribed Image Text: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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