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`._(92)^(238)U` has 92 protons and `238` nucleons. It decays by emitting an alpha particle and becomes:A. `._92^234 U`B. `._92^235 U`C. `._90^234 Th`D. `._93^237 Np` |
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Answer» Correct Answer - C ( c) Let the daughter nucleus be `._Z^A X`. So, reaction can be shown as `._92^238 U rarr ._Z^A X + ._2^4 He` From conservation of atomic mass `238 = A + 4` `rArr A = 234` From conservation of atomic number `92 = Z + 2` `rArr Z = 90` So, the resultant nucleus is `._90^234 X`, i.e., `._90^234 Th`. |
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