Question

The mean concentration of potassium in crustal rocks is 27 g/kg. If K-40 constitutes 0.012 atom percent of potassium, what is the K-40 activity in 1 ton of rock? (T1/2 = 1.3 x 109 years)

Answer 1

Given that Crustal rock contains mean concentration of Potassium is 27g in 1 kG rock

mass of the rock given as 1 ton = 907.185 kg

mean concetration of the potassium = 907.185kg * 27g/kg

mean concetration of the potassium = 24494 g

Number of moles of potassium = mass of K/molar mass of K = 24494g/(39.0983g/mol)

Number of moles of Potassium = 626.4723 moles

we know that 1 mole = 6.02214*10^23

Number of Potassium atoms = 626.4723 mol * 6.02214*10^23 atoms/mole

Number of Potassium atoms = 3.7727*10^26 atoms

K-40 constituency in given Pottasium = 0.012 atom percent

K-40 constituency in given Pottasium = 0.012/100*3.7727*10^26 =

K-40 constituency in given Pottasium(N) = 4.5272*10^22 atoms

given that half life of K-40 = 1.3*10^9 years

we know that decay constant λ = ln 2/t1/2

λ = ln2/(1.3*10^9) = 5.332*10^-10 year^-1

λ = 5.332*10^-10*1/(365*24*60*60) s^-1 = 1.6908 * 10^-17 s^-1

we have the formula for activity(A) in Bq

A = λ*N

λ = decay constant in s^-1

N = number of undecayed nucleii then,

A = activity in bequerels

A = 1.6908 * 10^-17 * 4.5272*10^22

A = 7.655 * 10^5 Bq