Explanation Given: Half-life of the isotope 239 P u as 24100 and its amount after 10,000 years as 0.4 g Formula used: We can use the growth and decay model y = C e k t . Logarithmic property: ln ( e x ) = x Also, e ln ( x ) = x Calculation: Let us consider the given expression: y = C e k t …… (1) Let us take initial amount as C . t = 10000 for the amount after 10,000 years and also given that y = 0.4 . By putting t = 10000 and y = 0.4 equation (1), we get: y = C e k t 0.4 = C e k × 10000 0.4 e 10000 k = C e k × 10000 e 10000 k 0.4 e 10000 k = C Half-life means t = 24100 and y = 1 2 C . By putting t = 24100 and y = 1 2 C in equation (1), we get: y = C e k t 1 2 C = C e k × 24100 1 2 = e 24100 k ln ( 1 2 ) = ln ( e 24100 k ) By simplifying further, 24100 k = ln ( 1 2 ) 24100 k = − 0

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ISBN: 9781285057095

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 6.2, Problem 36E

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To calculate: The initial amount of the radioactive isotope 239Pu and its amount after 1000 years.

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Chapter 6.2, Problem 35E

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The initial amount of the radioactive isotope 239 P u and its amount after 1000 years.

Solution Summary: The author calculates the initial amount of the radioactive isotope 239Pu and its amount after 1000 years.

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