Question 38296
you need to know the halflife of C-14. It is 5730 years.


Formula for half-life, T is {{{ T = (ln(2))/(y) }}} where y is the "time constant".


We need to figure out the time constant first:
{{{ 5730 = (ln(2))/y }}}
--> {{{ y = (ln(2))/(5730) }}}
y = 0.000120968 years^(-1)


Right then: exponential decay has a generic formula:
{{{ N = N[0]*e^(-yt) }}} 
where N is the amount left "now" and {{{ N[0] }}} is the original amount.


{{{ N/N[0] = e^(-yt) }}} 


Now, we are told that the carbon14 has reduced by 62%. So, if there was originally 100g of it, there now is 38g... a reduction of 62%. So, we have:


{{{ (38/100) = e^(-yt) }}} 
{{{ 0.38 = e^(-yt) }}}
{{{ ln(0.38) = -yt }}}
{{{ t = (ln(0.38))/(-y) }}}


--> {{{ t = (ln(0.38))/(-0.000120968) }}}


Working this out gives you 7998.7 years, which is in agreement with your answer, assuming rounding errors and the value of the half life used by the question.


jon.