1.

An element ._(96)X^(227) emits 4alpha and 5 beta particles to form new element Y. Then atomic number and mass number of Y are

Answer»

93, 211
211,93
212,88
88,212

Solution :`._(96)X^(227) rarr Y + 4 alpha + 5 BETA`
On equations mass number
`227 = y + 4 xx 4 + 0, y = 211`
On EQUATING atomic number
`96 + y + 2 xx 4 - 5, y = 93`


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