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Question

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
92U238zXA+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

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