To find:
(a) the number that goes into the elevated box (superscript) in the given nuclear reaction.
(b) the number that goes into the descended box (value of Z) in the given nuclear reaction.
Answer to Problem 1Q
Solution:
(a) The number that goes into the elevated box (superscript) in the given nuclear reaction = 42
(b) The number that goes into the descended box (value of Z) in the given nuclear reaction = 101
Explanation of Solution
Given:
i. A nuclear reaction 235U + n 132Sn +
ii. Refer to Appendix F and G of the textbook to know about A and Z of elements.
Concept
In any nuclear reaction, A and Z has to be conserved.
In the given reaction, balance A and Z on both sides of the reaction, because these numbers of conserved.
Let us say that the unknown nucleus ‘unknown’ has A=X and Z = Y
235U + n 132Sn +
(a)
On the left-hand side:
A = 235+1 = 236 ………………………………………………(1)
Z = 92 +0 = 92 …………………………………………………(2)
On the right-hand side:
A = 132 + X+ 3 = 135 + X ……………………………………..(3)
Z = 50 + Y + 0 = 50+ Y ………………………………………..(4)
Comparing eqn (1) and eqn (3) we can calculate that
X = 236 – 135 = 101
(b)
Comparing eqn (2) and (4) we can calculate that
Y = 91 – 50 = 42
From Appendix F and G, we can say that the ‘unknown’ element is Molybdenum Mo.
We can now write the complete nuclear reaction as
235U + n 132Sn +
Conclusion
We can complete a given nuclear reaction by calculating A and Z of elements on both the sides of the reaction.
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Chapter 43 Solutions
FUNDAMENTALS OF PHYSICS V.1
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