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S : When n/p ratio is high, nuclei emits...

S : When n/p ratio is high, nuclei emits `beta-` particles
E : The nuclear change is : `nto p""^(+)+e""^(-)+overset(-)v`

A

Both S and E are correct and E is cporrect explaination of S

B

Both S and E are correct and E not correct explaination of S

C

S is correct but E is wrong

D

S is wrong but the E is correct

Text Solution

Verified by Experts

The correct Answer is:
a
`N//A`
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Unstable nuclei attain stability through disintegration. The nuclear stability is related to neutron proton ratio (n//p) . For stable nuclei n//p ratio lies close to unity for elements with low atmoic numbers (20 or less) but it is more than 1 for nuclei having higher atomic numbers. Nuclei having n//p ratio either very high or low undergo nuclear transformation. When n//p ratio is higher than required for stability, the nuclei have the tendency to emit beta -rays. while when n//p ratio is lower than required for stability, the nuclei either emits alpha -particles or a positron or capture K -electron. beta -particle is emitted in radioactivity by

Unstable nuclei attain stability through disintegration. The nuclear stability is related to neutron proton ratio (n//p) . For stable nuclei n//p ratio lies close to unity for elements with low atmoic numbers (20 or less) but it is more than 1 for nuclei having higher atomic numbers. Nuclei having n//p ratio either very high or low undergo nuclear transformation. When n//p ratio is higher than required for stability, the nuclei have the tendency to emit beta -rays. while when n//p ratio is lower than required for stability, the nuclei either emits alpha -particles or a positron or capture K -electron. For reaction ._(92)M^(238) rarr ._(y)N^(x) + 2 ._(2)He^(4), ._(y)N^(x) rarr ._(B)L^(A) + 2 ._(-1)e^(0) The number of neutrons in the element L is

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