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Q.

Consider two nuclides X 88 z1  and Y 56 z2   The number of neutrons in X is 20 more than the neutrons in Y. Z1 and Z2 respectively, 

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a

37 and 26

b

39 and 26 

c

38 and 26 

d

38 and 25 

answer is C.

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Detailed Solution

 number of neutrons in x88=88−Z1
 number of neutrons in Y56=56−Z2
 given 88−Z1=56−Z2+20
Z2−z1=12

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