Answer
1.14 $\times$ $10^{20}$ no. of alpha emission will occur in 6.0 minutes.
Work Step by Step
Half life of Po-218 given = 3.0 minutes
That means, in 3.0 minutes half of initial amount of Po left.
Here, Initial amount = 55 mg
Amount of Po left in 3.0 minutes = $\frac{55}{2}$ mg
Hence, Amount of Po left in 6.0 minutes = $\frac{55}{4}$ = 13.75 mg
Therefore, Amount of Po that has been disintegrated in 6.0 minutes = (55-13.75) = 41.25 mg = 41.25 $\times$ $10^{-3}$ g
Now, Atomic mass of Po-218 = 218 amu (approx)
1 mole Po = 218 g Po
Hence, 218 g Po contains 6.023 $\times$ $10^{23}$ no. of Po atoms
Therefore, 41.25 $\times$ $10^{-3}$ g Po will contain $\frac{6.023 \times10^{23}\times41.25 \times10^{-3} }{218}$ $\approx$ 1.14 $\times$ $10^{20}$ no. of Po which will be disintegrated and emit same no. of alpha particle.
Answer: 1.14 $\times$ $10^{20}$ alpha emission will occur in 6.0 minutes.
