Radioactive decay involves the emission of n aplha particles and m beta particles. Given that the decay process involves the reaction 714N → n24He +m0-1β + n1n2X. Given that the values of n1,n2 , n and m are calculated theoretically such that n1 - m(d/dy (1/y)) = 7 where y=-1.
A. n2 = 14-n
B. n1 = 7
C. n2 =14-3n
D. m=2n
E. n1 = 7-2m + n
1/2 = e –(λ×T1/2)
C. n2 =14-3n
Comments
Leave a comment