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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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