For an extinct radionuclide, the slope of a correlation line in an isochron diag

Question

For an extinct radionuclide, the slope of a correlation line in an isochron diagram for D*/D vs P/D was said to equal (P*/P), where: P* = the amount of short-lived parent isotope P = the amount of stable parent isotope D* = the amount of daughter isotope produced by radioactive decay D = the amount of daughter isotope not involved in decay. It was also said that this slope was related to age.

(a) For a generic short-lived decay scheme, write an expression that relates the (P*/P) ratio to the time elapsed (t), assuming that the amount of radiogenic parent initially is given by P*o.

(b) Use your expression from part a to calculate the time elapsed between the formation of two objects in meteorites in which the decay of now-extinct 26Al appears to have occurred, one of which has (26Al*/27Al) = 5 x 10-5 (object A), the other of which has (26Al*/27Al) = 0.3 x 10-5 (object B). Round your answer to 2 significant digits. (Objects A and B are objects found in meteorites that appear to have pre-dated planetary formation.) HINT: Assume that object A gives the initial P*/P ratio. Decay constant for 26Al = 9.80 x 10-7 yr-1.

Explanation / Answer

a) from the isochron diagram we can see that p*= amount of parent isotope a,P=amount of stable parent isotope

now the unstable part is tansforming into daughter part which is based of D* and D the decay involved and not involved parts.

we consider the present time/sample time t such that P*(t)= P*0exp(-lamda*t)

where lamda=decay constant

So for daughter isotope D' = D+P*(t) exp(lamda*t-1)

where the second part is D* which is involved into decay process

therefore P*/P =exp(-lamda*t))

b) now for 1st case P*/P= 5*10^-5, lamda=9.8*10^-7

using the same we get 5*10^-5=exp(-9.8*10^-7t)

or time elapsed t= 1.01*10^7yrs

for second case P*/P= .3*10^-5, lamda=9.8*10^-7

therefore .3*10^-5=exp(-9.8*10^-7t)

or t= 1.29*10^7yrs

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