A radioactive nucleus undergoes a series of decays according to the sequence Aβ→A1α→A2α→A3 If the mass number and atomic number of A3 are 172 and 69respectively, then the mass number and atomic number of A is
56,23
180,72
120,52
84,38
A
180,72
B
56,23
C
84,38
D
120,52
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Solution
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AZA→A1→A2→69A1723
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