a fragment of a bone is discovered to 20% of the usual C-14 concentration.estimate the age of the bone,given that half life time of C-14 is5730years

Answer 1

The half life tells us that after 5730 years, half the original substance remains:

( 1 / 2 ) a = a * e ^ ( r * 5730 ) Divide both sides by a

1 / 2 = e ^ ( r * 5730 ) Take the natural logarithm of both sides

ln ( 1 / 2 ) = r * 5730 Divide both sides by 5730

ln ( 1 / 2 ) / 5730 = r

r = ln ( 1 / 2 ) / 5730

ln ( 1 / 2 ) = - ln 2

so:

r = - ln ( 2 ) / 5730

r = - 0.6931471806 / 5730

r = - 0.0001201

The decay will follow the equation:

Q(t) = a * e ^ (− 0.000121 t )

To find how old the bone fragment is that contains 20% of the original amount, we solve for t when Q(t) = 0.20 a

0.2 a = a * e ^ (− 0.000121 t ) Divide both sides by a

0.2 = e ^ (− 0.000121 t ) Take the natural logarithm of both sides

ln ( 0.2 ) = − 0.000121 t Divide both sides by − 0.000121

ln ( 0.2 ) / − 0.000121 = t

t = ln ( 0.2 ) / − 0.000121

t = -1.6094379124341 / − 0.000121

t = 13301.1397722 yrs

approx. 13 300 yrs