Question

Bismuth-210 is beta emitter with a half-life of 5.0 days.

Part A

If a sample contains 1.3 g of Bi-210 (atomic mass = 209.984105 amu), how many beta emissions occur in 14.0 days? Express your answer using two significant figures. betadecays

Part B

If a person's body intercepts 5.5 % of those emissions, to what dose of radiation (in Ci) is the person exposed? Express your answer using two significant figures. Dose of radiation = Ci

Answer 1

Moles of Bi = mass of Bi / atomic mass of Bi = 1.3 /209.984105 = 0.006191

Number of Bi atoms = 6.023 x 10^23 x moles = 6.023x10^23 x0.006191 = 3.7288 x 10^21

half life= 5 days

k = 0.693/ t1/2 =   0.693/5 = 0.13863   is decay constant

now t = 14   , we have formula t = ( 1/k) ln ( a/a-x)

where a= initial amount , a-x= final amount

14 = ( 0.13863) ln ( 3.7288 x 10^21 / a-x)

a-x = 0.5354 x10^21

Bi decayed = a-(-ax) = ( 3.7288-0.5354) x 10^21 = 3.2 x 10^21

Thus number of decays = 3.2 x 10^21

B) 5.5 % of emissions = ( 5.5/100) x 3.2 x 10^21 = 0.176 x 10^21

1 Ci = 3.7 x 10^10

hence dose in Ci = ( 0.176x10^21 /3.7x10^10) = 4.8 x 10^9