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Q.

A radioactive nucleus A decays to C after emitting two α and three β particles. Another nucleus B decays to C by emitting one α and five β particles. Half lives of A & B are 1hr and 2hr respectively and intermediaries have negligible half lives. At time t = 0; population of A is 4N0, of B is N0 and of C is zero.

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a

Atomic number of A – Atomic number of B=–2

b

When A & B have equal population, population of C=9N02

c

The decay rate of A & B are equal at t = 6hr

d

Atomic mass number of A – Atomic mass number of B = 4

answer is A, C, D.

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Detailed Solution

 Z1AA1Z11CA18 and Z2BA2Z2+3CA24

Z11=Z2+3Z1Z2=4

And A1- 8 = A2 – 4 A1A2=4

At time t NA=4N0eλ1tandNB=N0eλ2t When NA=NB4eλ1t=eλ2tt=2ln2λ1λ2=4hr

NC=4N0(1eλ1t)+N0(1eλ2t)

Putting the value to t = 4hr, we get NC=9N02

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