JEE ADVANCED 2016 SOLUTION - A plot of the number of neutrons N against the number of protons P

A plot of the number of neutrons (N) against the number of protons (P) of stable nuclei exhibits upward deviation from linearity for atomic number, Z 20. For an unstable nucleus having N/P ratio less than 1,
the possible mode(s) of decay is(are)
(A)  -decay (  emission) (B) orbital or K-electron capture
(C) neutron emission (D)  -decay (positron emission)
Sol. (B, D)
n 1
p

Positron, emission
b 0 b
a 1 a 1 X e Y    
 K - electron capture
b 0 b
a 1 a 1 X e Y    
Both process cause an increase in n/p ratio towards 1 thus stabilising the nucleus.


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