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Q.

When 92U238 decays it emits an α-particle. The new nuclide in turn emits a β-particle to give another nuclide X, The mass number and atomic number of X are, respectively

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a

234 and 91

b

234 and 96

c

232 and 88

d

234 and 88

answer is A.

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Detailed Solution

92U238⟶zXA+2He4+−1e0Equating mass number on both sides238 = A+4+0.'. A = 238 - 4 = 234 Equating atomic number on both sides92 = Z + 2 - 1:. Z = 92- 1 = 91

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