A wooden article found in a cave has only $40$ % as much $^{14} C$ activity as a fresh piece of wood. How old is the article? ( $t_{1 / 2}$ for $^{14} C = 5760$ years)

Open in App

Solution

$A_{0} = 100$
$A = 40$
$t = \frac{2.303 \times 5760}{0.693} \times l o g (\frac{100}{40})$
$t = 7617.29$ years

Suggest Corrections

Similar questions

Q. A piece of wood collected from Cro Magnon caves gave

4 d p m

. A freshly cut wood of the same weight gives

16 d p m

. The cro magnon man lived about

- ------- -

years ago
(Half life of C

^{14}

5760

years. Assume the activity is due to C

^{14}

only)

Q. Wooden article and freshly cut tree show activity of 7.6 and 15.2

m i n^{- 1} g m^{- 1}

of carbon (

t_{1 / 2} =

5760 years ) respectively. The age of article in years is:

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.