Half-Life and Radioactive Decay?

Question

The number of nuclei decaying in a sample of Gallium-68 decreased by a factor of 4 over a period of 136.6 minutes. What is the half life of this isotope?

Answer 1

The mass should satisfy the equation

m(t)=m0exp(-kt), where m is the mass, m0 represents the initial mass, and t is of course time. (This is found by solving the differential equation dm/dt=-kt, m(0)=m0.)

You are given that m(136.6)=m0/4. Plug this in and you can solve for k. Note that it doesn't matter whether or not we know exactly what m0 is, because it just cancels out when solving for k.

Once you have solved for k, to find the half life: you want to find the time t such that m(t)=m0/2, i.e. how long until only half of the original amount remains, so just set m(t)=m0/2 and solve for t.

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Just read Squeezie's answer...this is actually correct, but not the best approach in general. For instance, this logic would not work if you were told that it decreased by a factor of 6 in x minutes

Answer 2

well a factor of 4 means 2 times 2 or 2 half lives. So one half life is 136.6/2 or 68.3 minutes.

Answer 3

T a million/2= 0.693/lambda the position T1/2 = 0.5 existence, lambda = decay consistent ....................a million) for this reason, from the given expression and from a million) above, we've, Decay consistent ,Lambda = .693/2.062 = .336 in line with 12 months or, Lambda = .336 in line with 12 months = .028/month In sixty seven months = .028 *sixty seven= a million.876 will stay, i imagine.