If 128mg of U-238 was presented initially, how many half-lives would have to pas

Question

If 128mg of U-238 was presented initially, how many half-lives would have to pass until only 8.00mg of U-238 remained?
(Also the same question, but how many YEARS would have to pass?)
Can I please have the working out too as I would like to learn how to work it out on my own. Thanks :) If 128mg of U-238 was presented initially, how many half-lives would have to pass until only 8.00mg of U-238 remained?
(Also the same question, but how many YEARS would have to pass?)
Can I please have the working out too as I would like to learn how to work it out on my own. Thanks :)
(Also the same question, but how many YEARS would have to pass?)
Can I please have the working out too as I would like to learn how to work it out on my own. Thanks :)

Explanation / Answer

All radioactive decay follows first order kinetics.

Thus the half life is independent of inital concentration

method 1

128 mg ----------> 64mg -------------> 32 mg -----------> 16mg -----------> 8 mg

first 2nd 3rd 4th half lives

Thus it takes 4 half lives to reduce 128mg to 8 mg of U-238.

method 2

[C]t = [C]0 / 2n  where n is the number of half lives

8mg = 128mg / 2n or 2n = 16 , thus n = 4

part B

hlaf life of U-238 is 4.5 billion years = 4.5 x109 yrs

The number of years to make 128mg of U-238 to 8 mg = 4 x 4.5 x109 yrs

= 1.8x1010 yrs

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