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Problem 55 Hard Difficulty

In a piece of rock from the Moon, the $^{87} \mathrm{Rb}$ content is assayed to be $1.82 \times 10^{10}$ atoms per gram of material and the $^{87}$ Sr content is found to be $1.07 \times 10^{9}$ atoms per gram. (The relevant decay is $^{87} \mathrm{Rb} \rightarrow^{87} \mathrm{Sr}+\mathrm{e}^{-} .$ The half-life of the decay is $4.8 \times 10^{10}$ yr.) (a) Determine the age of the rock. (b) Could the material in the rock actually be much older? (c) What assumption is implicit in using the radioactive -dating method?

Answer

a. 4.1 \times 10^{9} y r
b. \text { see explanation }
c. \text { see explanation }

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Video Transcript

number 55. We have rubidium 87 and it is decaying into astronomy 87. So it's a beta decay and the rubidium in this rock sample, there's this money, Adam's program of the Radium and this money, Adam's Program of Australia and The Half Life of the Rubidium. Is this? So we're looking for how old this piece of rock is, the question I was like to use here, his this one. But where N is the number of nuclei they end with, I mean the part that's still radioactive, we're and then the original An is how much time a nuclear you started with before the decay 1/2 raised to the and which is number of half. Life's just a easy equation of work with, Oh, so after this decay, this is still radioactive. So that's my ending mail. So 1.82 times 10 of attempt, How much I started with would be both of those together because originally there were all this rubidium, and then some of it decayed in the strong room. So I'm adding these two numbers together to get this and and that's 1.93 times 10 to the 10th. This is 1/2 when, Right, That doesn't desperately guess Raised to the end, you know, some flies a little bit, so I'm gonna solve for that end. So I'm going to divide 1.82 by 1.83 and I get 0.8 943 And this would be times 0.5 raise to the m to solve something. Because when we take the log of both sides excuse me, someone have the log. A 0.9 for three equals a log of 0.5 raised to the end. But remember how this works. The exponents you can bring that down is a coefficient in front. She was the end times the wog 0.5. So now if I just take the log of this divided by the log of that I would get in for N I'm getting point. Oh, eat 47 Remember, end is that's the number of half lifes I have. So that's not the answer I have. You know, this fraction of 1/2 life and this is my half life. So can I just take this times? That would be how old it is. Some taking 0.847 times, 4.8 times 10 to the 10th. And I'm getting four point. Oh, seven times 10 to the knife. And that will be in years. Then it goes on to ask is could have actually older than that. And no, it can't. It can't be over that, because the assumption we made here was that all of the strong Thiam used to be rubidium. So this original number that we started with with some of both of those, but in fact, some of the rock the rock could have had some strong to begin with, so it no, it can't be any older than this. It could be younger, but it can't be over. Um, just read that. Cannot the older the assumption we made, um So our assumption is all of the strontium waas rubidium

University of Virginia
Top Physics 103 Educators
Elyse G.

Cornell University

LB
Liev B.

Numerade Educator

Aspen F.

University of Sheffield

Meghan M.

McMaster University