Radioactive Decay: Finding the Half-Life of K-40 in KCL

[JingXionG]

A 2.71g sample of KCL is found to be radioactive, and it is decaying at a constant rate of 4490 Bq. The decays are traced to the ekement potassium and in particular to the isotope K(proton number 40), which constitutes 1.17% of normal potassium. find the half-life of this nuclide( Take molar mass of KCL = 74.555g)

Ans provided is 1.25 x 10^9 years. I can't seem to even come close to that value.

I found the number of K-40 in the sample which is 2.560 x 10^20. From there, using A= constant x number of particles of K-40 and rate of decay = -dN/dt, I can't seem to get that ans

 

[Grogs]

I think you're on the right track here. Your number for the number of ⁴⁰K atoms is very close to what I got.

My next step was to find the decay constant, [lambda]. If the rate of decay is constant, then

[lambda] = decays per second / number of atoms

4490 Bq = 4490 decays per second. Once you have [lambda], you can use

T_1/2 = ln(2) / [lambda] to get your half-life in seconds. I get 3.95 x 10 ¹⁶ seconds, which is equivalent to 1.25 Billion years.

I should add a word of caution that this is *not* a completely rigorous method. You can't use it for something with a half-life of, say, 15 seconds. However, since the problem states the decay rate is constant and you can see that the decays/second are << than the number of atoms in the sample (16 orders of magnitude in this case), it should get an accurate answer.

 

[JingXionG]

argh

i know where i went wrong, forgot to convert the seconds to years!