Explanation Given Information: The provided half-life of C 14 is 5715 years and the initial quantity is 3 grams. Formula used: The equation for a positive quantity y whose rate of change with respect to time is proportional to the amount of quantity is, y = C e k t Where, C is the amount of quantity, t is the any time and k is the proportionality constant. The exponential growth when k > 0 and the exponential growth when k < 0 . The property of natural logarithm, ln x n = n ln x Calculation: Consider the provided information that the half-life of C 14 is 5715 years and the initial quantity is 3 grams. Now, at t = 0 , the initial amount y = 3 . Therefore, the value of C is 3 . Substitute C = 3 in the equation y = C e k t . y = 3 e k t Also the half-life of C 14 is 5715 years, that is, at t = 5715 the initial amount reduces to half so, y = 3 2 = 1.5 Substitute t = 5715 and y = 1.5 in the equation y = 3 e k t . 1.5 = 3 e k ⋅ ( 5715 ) e 5715 k = 1.5 3 Take natural log on both the sides and use the property of logarithm, ln x n = n ln x . ln ( e 5715 k ) = ln ( 1.5 3 ) 5715 k ⋅ ( ln e ) = ln ( 1 2 ) 5715 k = ln ( 1 2 ) k = ln ( 1 2 ) 5715 Therefore, the value of k is ln ( 1 2 ) 5715

CALCULUS APPLIED APPROACH >PRINT UGRADE<

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ISBN: 9780357667231

Author: Larson

Publisher: CENGAGE L

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Chapter 4.6, Problem 12E

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To calculate: The amount of C14 left after 1000 years and 10,000 years, if the half-life of C14 is 5715 years and the initial quantity is 3 grams.

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Chapter 4.6, Problem 11E

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CALCULUS APPLIED APPROACH >PRINT UGRADE<

Ch. 4.1 - Use the properties of exponents to simplify each...Ch. 4.1 - Use the properties of exponents to simplify each...Ch. 4.1 - Use the formula for the ratio of carbon isotopes...Ch. 4.1 - Prob. 4CP Ch. 4.1 - Prob. 5CP Ch. 4.1 - Prob. 1SWU Ch. 4.1 - Prob. 2SWU Ch. 4.1 - Prob. 3SWU Ch. 4.1 - Prob. 4SWU Ch. 4.1 - Prob. 5SWU

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The amount of C 14 left after 1000 years and 10 , 000 years, if the half-life of C 14 is 5715 years and the initial quantity is 3 grams.

Solution Summary: The author calculates the amount of 14C left after a half-life of 5715 years and the initial quantity is 3 grams.

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To calculate: The amount of C14 left after 1000 years and 10,000 years, if the half-life of C14 is 5715 years and the initial quantity is 3 grams.

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