Question

half life of carbon-14 is 5730 years. all radioactive decay is first order. measurments taken indicate that scrolls contain 79.5% of carbon-14. how old are these scrolls?


I derived the rate law to get half life so I could solve for k. but now I don't know what to do

Answer 1

\[N(t)=N_0e^{-\lambda t}\\
\frac{N(t)}{N_0}=e^{-\lambda t}\\
t =-\frac1\lambda\ln\frac{N(t)}{N_0}\]

\[\lambda = 1/\tau\\\qquad\qquad\qquad\qquad\tau=\frac{t_{1/2}}{\ln 2}\\
\lambda=\frac{\ln 2}{t_{1/2}}\]

\[t=\frac{-t_{1/2}}{\ln 2}\ln\frac{N(t)}{N_0}\]

Answer 2

You can use the equation

y = A * (1/2) ^ (t/5730)

where A is the initial amount of carbon-14 in grams
and y is the amount remaining after t years

We are given that there 79.5% of the initial amount remaining.

.795*A = A * ( 1/2)^(t / 5730)

you can divide both sides by A

.795 = ( 1/2) ^( t / 5730)

solve for t

ln( .795) = ln( 1/2) ^( t / 5730)

ln(.795) = t/5730 * ln(1/2)

ln(.795)/ln(1/2) = t/5730

5730 * ln(.795)/ln(1/2) = t

Answer 3

Can you solve it now?