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

t = 4254 y e a r s

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t = 1254 y e a r s

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t = 2254 y e a r s

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None of these

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Solution

The correct option is A $t = 4254 y e a r s$
$λ = \frac{2.303}{t} l o g \frac{C_{0}}{C}$
$\frac{0.693}{t_{50}} = \frac{2.303}{t} l o g \frac{C_{0}}{C}$
$\frac{0.693}{5770} = \frac{2.303}{t} l o g \frac{100}{60}$
$t = 4254 y e a r s$

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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.

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A piece of wood from an archaeological source shows a 14C activity is 60% of the activity found in fresh wood today. Calculate the age of the archaeological sample. (t1/2=C=5770yr)

The correct option is A t=4254years λ=2.303tlogC0C0.693t50=2.303tlogC0C0.6935770=2.303tlog10060 t=4254years

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

The correct option is A $t = 4254 y e a r s$
$λ = \frac{2.303}{t} l o g \frac{C_{0}}{C}$
$\frac{0.693}{t_{50}} = \frac{2.303}{t} l o g \frac{C_{0}}{C}$
$\frac{0.693}{5770} = \frac{2.303}{t} l o g \frac{100}{60}$
$t = 4254 y e a r s$