Ratio of the rate of decay of R to that of S after 2 hours

[moenste]

Homework Statement

At a certain time, two radioactive sources R and S contain the same number of radioactive nuclei. The half-life is 2 hours for R and 1 hour for S. Calculate:

(a) the ratio of the rate of decay of R to that of S at this time.
(b) the ratio of the rate of decay of R to that of S after 2 hours.
(c) the proportion of the radioactive nuclei in S which have decayed in 2 hours.

Answers: (a) 1 : 2, (b) 1 : 1, (c) 75 %.

2. The attempt at a solution
I get how to approach (c). At t = 0 hours we have 100 % of S, in 1 hour we'll have 50 % of it and in one more hour we'll have half of that -- 25 %. So in two hours 75 % of S decayed.

Though I don't quite understand what is required in (a) and (b). In both cases they decay 1 : 2, since S decays faster. Maybe I miss something.

 

[Jonathan Scott]

Note that the rate of decay is proportional to the remaining number of nuclei.

 

[moenste]

Note that the rate of decay is proportional to the remaining number of nuclei.

(a) t = 0. dN / dt = ?
N_R = 100 % = N_S.
T_{1/2 R} = 2 hours
T_{1/2 S} = 1 hour.
dN / dt = - λ N
dN / dt = - (ln 2 / T_1/2) * N

R [dN / dt] = 9.6 * 10^-5
S [dN / dt] = 1.9 * 10^-4

R for S = 0.5 or 1 : 2.

(b) t = 2 hours
N_R = 50 %, N_S = 25 %.
T_{1/2 R} = 2 hours
T_{1/2 S} = 1 hour.
dN / dt = - λ N
dN / dt = - (ln 2 / T_1/2) * N

R [dN / dt] = 9.6 * 10^-5 * 0.5 = 4.8 * 10^-5
S [dN / dt] = 1.9 * 10^-4 * 0.25 = 4.75 * 10^-5

R for S = 1 : 1.

Like this, right?

 

[Jonathan Scott]

Probably (I haven't checked carefully) but that's far more complicated than you need, as you can simply use ratios between the two cases rather than working them out. As you've already used above, exponential decay rates vary as the inverse of the half life, so if the half lives are in the ratio m:n then the decay rates for the same number of nuclei are in the ratio n:m. Then after 2 hours the amount of R is 1/2 of the original and the amount of S is 1/4 of the original, so there's now twice as much R left as S.