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
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
1 answer