Question

The fraction of a radioactive isotope remaining at time t is (1/2)^t/t1/2 where t1/2 is the half-life. If the half-life of carbon−14 is 5,730 yr, what fraction of carbon−14 in a piece of charcoal remains after
(a) 14.0 yr?

(b) 1.900 × 10^4 yr? _________ × 10 (Enter your answer in scientific notation.)

(c) 1.0000 × 10^5 yr? __________× 10 (Enter your answer in scientific notation.)

Answer 1

Answer:

a) $1.065\times 10^{-8}$fraction of carbon−14 in a piece of charcoal remains after 14.0 years.

b) $0.000\times 10^{-3}$fraction of carbon−14 in a piece of charcoal remains after  $1.900\times 10^4 years$

c) $0.0000\times 10^{-4}$fraction of carbon−14 in a piece of charcoal remains after  $1.0000\times 10^5 years$.

Explanation:

The fraction of a radioactive isotope remaining at time t is given by:

$[A]=\frac{(\frac{1}{2})^t}{t_{\frac{1}{2}}}$

Taking log both sides:

$\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}]$

[A] =  fraction at given time t

$t_{\frac{1}{2}}$ = half life of the carbon−14 =5,730 years

a)When , t = 14 years

$\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}]$

$\log [A]= 14 years\times (-3010)-\log [5,730 years]$

$[A]=1.065\times 10^{-8}$

b)When , t = $1.900\times 10^4 years$

$\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}]$

$\log [A]= 1.900\times 10^4 years\times (-3010)-\log [5,730 years]$

$[A]=0.000\times 10^{-3} [/tex$

c)When , t = $1.0000\times 10^5 years$

$\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}]$

$\log [A]= 1.0000\times 10^5 years\times (-3010)-\log [5,730 years]$

$[A]=0.0000\times 10^{-4}$