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1. The half-life of 238 92U is 4.5 × 10

(ii) A certain time, t, after its formation the sample contained twice as many 238 92U atoms as 206, 82 Pb atoms. Show that the number of 238, 92 U atoms in the rock sample at time t was 2.0 × 10

The answer shows an equation u/(3 x10

(u represents the number of uranium atoms then)

How did they get this equation? What kind of way of thinking got them to this point... don't get it. If someone could explain please. Also, if it helps, there was also part (i) where the task was to show how the number of 238 92U atoms and the number of 206 82Pb atoms in the rock sample vary over a period of 1.0 × 10

Maybe I should add more to be up with the forum rules... I want to use N=N

Question 2 in the paper: http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN12.PDF

Mark scheme: http://filestore.aqa.org.uk/subjects/AQA-PHYA52A-W-MS-JUN12.PDF

^{9}years, which is much larger than all the other half-lives of the decays in the series. A rock sample when formed originally contained 3.0 × 10^{22 }atoms of 238 92U and no 206 82Pb atoms. At any given time most of the atoms are either 238 92U or 206 82Pb with a negligible number of atoms in other forms in the decay series.(ii) A certain time, t, after its formation the sample contained twice as many 238 92U atoms as 206, 82 Pb atoms. Show that the number of 238, 92 U atoms in the rock sample at time t was 2.0 × 10

^{22}.The answer shows an equation u/(3 x10

^{22}- u) = 2(u represents the number of uranium atoms then)

How did they get this equation? What kind of way of thinking got them to this point... don't get it. If someone could explain please. Also, if it helps, there was also part (i) where the task was to show how the number of 238 92U atoms and the number of 206 82Pb atoms in the rock sample vary over a period of 1.0 × 10

^{10}years from its formation using a graph of N against time t.Maybe I should add more to be up with the forum rules... I want to use N=N

_{0}e^{-yt}and worked out the decay constant y to be ln(2)/Half life of U = ln(2)/4.5x10^{9}= 1.54x10^{-10}.Question 2 in the paper: http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN12.PDF

Mark scheme: http://filestore.aqa.org.uk/subjects/AQA-PHYA52A-W-MS-JUN12.PDF

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