# Modern Physics - Radioactivity

Please help with the following problem.

The number of unstable nuclei remaining after a time t=5.00 yr is N, and the number present initially is N 0. Find the ratio N/N0 for (a) ^14/6C (half-life=5730 yr), (b) ^15/8O (half-life=122.2x; use t=1.00h, since otherwise the answer is out of the range of your calculator), and (c) ^3/1H (half-life=12.33 yr).

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Problem:The number of unstable nuclei remaining after a time t=5.00 yr is N, and the number present initially is N 0. Find the ratio N/N0 for (a) ^14/6C (half-life=5730 yr), (b) ^15/8O (half-life=122.2x; use t=1.00h, since otherwise the answer is out of the range of your calculator), and (c) ^3/1H (half-life=12.33 yr).

The radioactive decay model is N = N0.e^ -kt. So, N/N0 = e^ -kt.

(a) For C14/6, half-life = 5730 years. So, 0.5 = e^ -k(5730). Solving, we get,

k = 1.21 x 10^ -4. Hence, N/N0 = e^ -(1.21 x 10^ -4)t.

At t = 5, N/N0 = e^ -(1.21 x 10^ -4) x 5 = 0.9994.

(b) For O15/8, half-life = 122.2 s. So, 0.5 = e^ -k(122.2). Solving, we get,

k = 0.0057. Hence, N/N0 = e^ -(0.0057)t.

At t = 1 hour = 3600 s, N/N0 = e^ -(0.0057) x 3600 = 1.35 x 10^ -9.

(c) For H3/1, half-life = 12.33 years. So, 0.5 = e^ -k(12.33). Solving, we get,

k = 0.056. Hence, N/N0 = e^ -(0.056)t.

At t = 5, N/N0 = e^ -(0.056) x 5 = 0.755.

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