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(a) ²¹²₈₃Bi ²⁰⁸₈₁Ti +

²¹²₈₃Bi ²¹²₈₄Bi + ^-

(b) t_1/2= 1h

After 1h, ²¹²₈₃Bi will be half decayed

So, ²¹²₈₃Bi is 0.5g present

²⁰⁸₈₁Ti and ²¹²₈₄Bi will be formed in the ratio 7/13 and total mass of the sample should be 1g. So, total mass of ²⁰⁸₈₁Ti and ²¹²₈₄Bi is 0.5g.

⇒ Mass of ²⁰⁸₈₁Ti will be ratio of mass present × total mass=

And mass of ²¹²₈₄Bi will be of mass present × total mass=