Question

the sequence of stepwise decays of radioactive nucleus X( Z= 71,A= 176) would finally result a nucleus X2 in the process.

X (z=71,A=176) ------alpha----------- X1------beta------ X2; X2 is
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Answer 1

71=176=17-25=2581-29174=32919=1-

Answer 2

In alpha decay, the atomic number decreases by 2 and the mass number decrease by 4

So,

X (z=71,A=176) ⇒ X1 ( z = 69, A = 172)

In beta decay, the atomic number increases by 1, but the mass number remains the same

X1 ( z = 69, A = 172) ⇒ X2 ( ( z = 70, A = 172)

So, the final nucleus will have an atomic number of 70 and mass number of 172.