1.

In the sequence of the following nuclear reaction ""_(98)^(238) X overset(-alpha)(to) Y overset(-beta)(to) Z overset(-beta) (to)Loverset(n alpha) (to) ""_(90)^(218) M What is the value of n

Answer»

3
4
5
6

Solution :`""_(98) X^(238) OVERSET(- (""_(2) He^(4)))(to) ""_(96) Y^(234) overset(- (""_(1) beta^(0)))(to) ""_(97) Z^(234) overset(- (""_(1) beta^(0)))(to) ""_(98) Z^(234)`
`""_(98) Z^(234) to ""_(90) Na^(218) + N(""_(2) alpha^(4))`
`234 = 218 + 4N implies n = ((234 - 218)/(4)) = 4 , 98 = 90 + 2N implies n = (8)/(2) = 4`


Discussion

No Comment Found