Question

A sample of drinking water collected from a suburban Boston municipal water system in 2002 contained 0.55 pCi/L of radon. Assume that this level of radioactivity was due to the decay of 222_Rn (t_i/2 = 3.8 days). 1 picocurie (pCi) = 1 times 10^-12 Curies (a) What was the level of radioactivity of this nuclide in the sample? Number (b) How many disintegrations per hour would occur in 2.5 L of the water?

Answer 1

I will just anwer a question at the same time

(a) What was the level of radioactivity of this nuclide in the sample

You have

0.55 pCi/L you must to transform into Curies and then into Becquerel

The convertion factor is: 0.55 pCi = 0.55*10 ^-12 Ci

a) (0.55 *10 ^-12 Ci / L)*(3700.000.000 Becquerel/1 Curie)= 0.002035 Bq/L = 2.035 *10 ^-3 Bq/L

b) Now you have

t1/2= ln2*/ λ then

λ=ln2/t1/2=(ln 2 / 3.8 days )* (1 dia/24 hours)=0.0076 h ^-1

^{Now 1Bq= DFS( Desintegration for second)}

You have

2.035 *10 ^-3 Bq/L *2.5 L= 5.0875*10-³ Bq* 1DFS/ 1Bq =  5.0875*10-³ DFS*(3600 SEC/1 H)= 18.315 Desintegration/H.