Question

The naturally occurring isotope K40 is widely spread in the environment.

Infact, the average concentration of potassium in the crustal rocks is 27 [g/kg] and in the oceans is 380 [mg/liter].

K40 occurs in plants and animals, has a half-life of 1.3 billion years and an abundance of 0.0119 percent.

Potassium's concentration in humans is 1.7 [g/kg].

In urine, potassium's concentration is 1.5 [g/liter].

1) Calculate the specific activity of K40 in Becquerels per gram and in Curies/gm of K40.

2) Calculate the specific activity of K40 in Becquerels per gram and in Curies per gm of overall potassium.

3) Calculate the specific activity of K40 in urine in [Bq/liter].

Answer 1

Specific activity is defined as the amount of radioactivity - or the decay rate of a particular radionuclide per unit mass of the radionuclide

1)

The specific activity of K40 is denoted by the letter R

R = (ln2N_aAb) / (t_1/2A)

ln2 = 0.693

Avogadro's no (N_a) = 6.021023 atoms

40 K abundance (Ab) = 0.0119% = 0.0119/100 = 0.000119 nuclei

40 K half life (t_1/2) = 1.3 billion years = 1.310⁹365246060 s,   [convertion of billion years into seconds]

Atomic weight (A) = 39.1

Specific activity of K 40 = (0.6936.0210²³0.000119) / (1.310⁹36524360039.1)

Specific activity of K 40 = (4.964510^-410²³) / (160297488010⁹)

Specific activity of K 40 = (4.964510¹⁹) / (1.610⁹10⁹) = (4.964510¹⁹) / (1.610¹⁸)

Specific activity of K 40 = 31.028 Bq/g

1 Bq/g = 2.702702710^-11 curies/g [unit convertion]

Therefore 31.028 Bq/g = 31.0282.702702710^-11 curies/g = 83.85910^-11 curies/g

likewise we can easily find the remaning (2) and (3) parts of the problem