Answer
Approx. 1.72 $\times$ $10^{21}$ no. of beta emissions will occur in 5.0 days.
Work Step by Step
Half-life of Bi-210 = 5.0 days
That means In 5.0 days the amount of Bi will become half of the initial amount.
Initial amount = 1.2 g
Hence, the amount of Bi that will be disintegrated in 5.0 days= 1.2$\div$2 = 0.6 g.
Now, Atomic mass of Bi-210 = 210 amu.
hence, 1 mole Bi-210 = 210 g Bi.
We know that 1 mole of Bi-210 contains 6.023 $\times$ $10^{23}$ no. of Bi atoms (According to Avogadro's law)
Therefore, 210 g Bi-210 contains 6.023 $\times$ $10^{23}$ no. of Bi atom
So, 0.6 g Bi-210 contains 6.023 $\times$ $10^{23}$ $\times$ 0.6 $\div$210 = 1.72085714 $\times$ $10^{21}$ no. of Bi atom $\approx$ 1.72 $\times$ $10^{21}$ no. of Bi atom.
Therefore, approx. 1.72 $\times$ $10^{21}$ no. of beta emissions will occur in 5.0 days
