Answer to Question #182122 in Atomic and Nuclear Physics for AlexAlale

Question #182122

A radioisotope has an initial activity of 5 millicurie. 48hours later, the observed activity is 4 millicurie. Determine: a) The half life of the radioisotope

b) The initial number of nuclides in the radioisotope.

Nb,Show workings



1
Expert's answer
2021-04-19T17:10:44-0400

"\\displaystyle A(t) = A_0 2^{-\\frac{t}{T_{1\/2}}}"

a) "A_0= 5, A=4, t=48 h"

"\\displaystyle \\frac{A}{A_0} = 2^{-\\frac{t}{T}}"

"\\displaystyle \\frac{t}{T}= log_2{\\frac{A_0}{A}}=log_2 \\frac{5}{4} = 0.3219"

"\\displaystyle T = \\frac{48 h}{0.3219}=149.11 \\, h"


b) We will need to convert everything to SI units, curie -> Bq, h -> s.

"\\displaystyle A_0= N_0 \\frac{ln2}{T}"

"\\displaystyle N_0 = \\frac{A_0 T}{ln 2}=\\frac{5 \\cdot 10^{-3}\\cdot 3.7\u22c510^{10} \\cdot 149.11 \\cdot 3600}{0.6931} =14. 328 \\cdot 10^{13}= 1.4328 \\cdot 10^{14}"


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