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 :
A
234 and 91
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B
234 and 96
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C
232 and 88
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D
234 and 88
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Solution
The correct option is A 234 and 91 92U238→zXA+2He4+−1e0 Equating mass number of both sides 238 = A + 4 + 0 ∴ A = 238 - 4 =234 Equating atomic number of both sides 92 = Z + 2 - 1 ∴ Z = 92 - 1 = 91