A piece of wood from an archeological source shows a C-14 activity which is 60% of the activity found today. Calculate the age of the sample $(t_{\frac{1}{2}}$ for $_{6}^{14} C = 5770 y e a r s)$ :

3515 years

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300 years

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4253 years

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4000 years

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Solution

The correct option is C 4253 years
As we know,
Age = $2.303 t_{1 / 2} / 0.693 l o g (A_{0} / A)$
$= 2.303 * 5770 / 0.693 l o g (1 / 0.6) = 4253$ years

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Similar questions

Q. A piece of wood from an archaeological source shows a

^{14} C

activity is 60% of the activity found in fresh wood today. Calculate the age of the archaeological sample.

(t_{1 / 2} = C = 5770 y r)

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Q. A wooden article found in a cave has only

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Q. The amount of

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A piece of wood was burned and 7.32 g of $C O_{2}$ was collected. The total radioactivity in $C O_{2}$ was 7.6 dpm. How old was the wood sample ? Given $t_{\frac{1}{2}}$ of C - 14 = 5730 years, specific activity of carbon obtained from a fresh wood sample is 15.2 dpm.

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A piece of wood from an archeological source shows a C-14 activity which is 60% of the activity found today. Calculate the age of the sample (t12 for 146C=5770years) :

The correct option is C 4253 yearsAs we know,Age = 2.303t1/2/0.693log(A0/A)=2.303∗5770/0.693log(1/0.6)=4253 years

A piece of wood from an archeological source shows a C-14 activity which is 60% of the activity found today. Calculate the age of the sample $(t_{\frac{1}{2}}$ for $_{6}^{14} C = 5770 y e a r s)$ :

The correct option is C 4253 years
As we know,
Age = $2.303 t_{1 / 2} / 0.693 l o g (A_{0} / A)$
$= 2.303 * 5770 / 0.693 l o g (1 / 0.6) = 4253$ years