1.

The ""_(92)^(235) U disintegrates to give 4 alpha and 6 alpha particles . The atomic number of the end product is

Answer»

92
96
84
90

Solution :`""_(92) U^(235) to ""_(Z)X^(A) + 4 (""_(2) He^(4)) + 6 (""_(-1) BETA^(0)) , 235 = A + 16 + 0 implies A = 235 - 16 = 219`
`92 = Z + 8 - 6 implies Z = 92 - 2 = 90`


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