Answer to Question #146174 in General Chemistry for Cyryl Mayuga

Question #146174

A wooden Japanese temple guardian statue of the Kamakura period (AD 1185 - 1334) showed a carbon-14 activity of 12.9 dpm/g-C when it was discovered in 1990. If the initial 14C activity was 15.0 dpm/g-C, what year was the statue made? ( t½ = 5730 yrs)

Expert's answer

Radioactive decay follows first order kinetics.Therefore,Decay constant (λ)=t1/20.693=57300.693Given, [R0]=100∴[R]=80and t=λ
2.303log[R][R]0=(5730
0.693)
2.303log80100=0.693
2.303×5730×0.0969=1845 years.

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26.11.20, 21:02

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Answer to Question #146174 in General Chemistry for Cyryl Mayuga

Radioactive decay follows first order kinetics.Therefore,Decay constant (λ)=t1/20.693=57300.693Given, [R0]=100∴[R]=80and t=λ
2.303log[R][R]0=(5730
0.693)
2.303log80100=0.693
2.303×5730×0.0969=1845 years.

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Answer to Question #146174 in General Chemistry for Cyryl Mayuga

Radioactive decay follows first order kinetics.Therefore,Decay constant (λ)=t1/2​0.693=57300.693​Given, [R0​]=100∴[R]=80and t=λ2.303​log[R][R]0=(57300.693)2.303​log80100=0.6932.303×5730​×0.0969=1845 years.

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Related Questions

Radioactive decay follows first order kinetics.Therefore,Decay constant (λ)=t1/20.693=57300.693Given, [R0]=100∴[R]=80and t=λ
2.303log[R][R]0=(5730
0.693)
2.303log80100=0.693
2.303×5730×0.0969=1845 years.