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Standard IX

Chemistry

Radiocarbon Dating

A piece of wo...

Question

A piece of wood from an archaeological source shows a $^{14} C$ activity which is $60$ % of the activity found in fresh wood today. The age of archaeological sample will be:

4246

yrs

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4624

yrs

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4628

yrs

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6248

yrs

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Solution

The correct option is A $4246$ yrs
$k = \frac{0.693}{5760} {y e a r s}^{- 1} S o, 0.6 A_{0} = A_{0} e^{- \frac{0.693}{5760} t} - ln (0.6) = \frac{0.693}{5760} t t = 4246 y e r s .$

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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)

Q. A piece of wood from an archaeological sample has

5.0 c o u n t s {m i n}^{- 1}

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The half-life for radioactive decay of $^{14} C$ is 5730 yr. An archaeological artifact containing wood had only $80 %$ of the $^{14} C$ found in a living tree. Estimate the age of the sample.

Q. 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

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_{6}^{14} C = 5770 y e a r s)

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activity left in the piece?

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A piece of wood from an archaeological source shows a 14C activity which is 60% of the activity found in fresh wood today. The age of archaeological sample will be:

The correct option is A 4246 yrsk=0.6935760years−1So,0.6A0=A0e−0.6935760t−ln(0.6)=0.6935760tt=4246yers.

A piece of wood from an archaeological source shows a $^{14} C$ activity which is $60$ % of the activity found in fresh wood today. The age of archaeological sample will be:

The correct option is A $4246$ yrs
$k = \frac{0.693}{5760} {y e a r s}^{- 1} S o, 0.6 A_{0} = A_{0} e^{- \frac{0.693}{5760} t} - ln (0.6) = \frac{0.693}{5760} t t = 4246 y e r s .$