i got your half-life right here!?

Question

the half-life of radioactive carbon is 5600 years meaning that after every period of 5600 years only half of the radioactive carbon atoms remain. how long is it before a sample with ,10 to the power of six, radioactive carbon atoms decays enough so that fewer100 such atoms remain?

Answer 1

The decay is of the form N(t)/N0 = e^-t/h, where h is a constant. When t = HL (half life time) N(t)/N0 = .5 therefore N(HL)/N0 = .5 = e^-HL/h. Take the natural log of both sides to get ln(.5) = -HL/h, so h = HL/ln(2). Therefore the formula for decay is

N(t)/N0 = e^-t*ln(2)/HL

You want to find tx such that N(t)/N0 = 100/10^6 = 10^-4. That means 10^-4 = e^-tx*ln(2)/HL. Again, taking log (base 10) of both sides gives you -4 = log(e^-tx*ln(2)/HL) or[ -tx*ln(2)/HL]*ln(10). The time tx is then 4*HL*ln(10)/ln(2). This comes out 74,411years

Answer 2

If you have a piece of radioactive carbon and it has 10 to the power of 6 carbom atoms, it would take 11200 years for the substance to be completely depleted of radioactive carbon atoms. So a logical guess would be between 11196 to 11199 years before the overall radioactive atoms are less than 100

Answer 3

It's a logarithmic function. 10^6(1/2)^n=10^2 so (1/2)^n=10^-4. n is the number of half lives. You figure it out from there. I won't do your homework for you. Only an idiot would do that.