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The half-life of 234U, uranium-234, is 2.52 multiplied by 105 yr. If 98.6% of the uranium in the original sample is present, what length of time (to the nearest thousand years) has elapsed?
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The half-life of 234U, uranium-234, is 2.52 multiplied by 105 yr. If 98.7% of the uranium in the original sample is present,
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(1/2)^(T/2.52*10^5) = 0.988 T is the time, in years, since it was 100% U-234. Solve for T. Use of
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u = Uo e^-kt u/Uo = .5 = e^-k(2.52*10^5) -ln .5 = -2.53*10^5 k k =2.74*10^-6 .977 = e^-kt -ln .977 =
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The half-life of 234U, uranium-234, is 2.52 multiplied by 105 yr. If 98.3% of the uranium in the original sample is present,
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he half-life of 234U, uranium-234, is 2.52 multiplied by 105 yr. If 97.5% of the uranium in the original sample is present, what
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the fraction after t years is (1/2)^(t/252000) so, solve for t in .975 = (1/2)^(t/252000) t = 9204
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see this http://www.jiskha.com/display.cgi?id=1311013577 just change the numbers.
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The half-life of 234U, uranium-234, is 2.52 105 yr. If 98.6% of the uranium in the original sample is present, what length of
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.986=e^(-.693t/2.52E5) take the ln of both sides.. > -0.0140989244=-.693t/2.52E5 solve for t t==
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The half-life of 234U, uranium-234, is 2.52 105 yr. If 97.7% of the uranium in the original sample is present, what length of
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.5 = e^2.52*10^5 k ln .5 = 2.52 * 10^5 k k =-2.75*10^-6 ln .977 = -2.75*10^-6 t t = 8460 years =
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