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 areA. 93, 211B. 211,93C. 212,88D. 88,212

Answer» Correct Answer - A
`._(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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