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

a)
`1.065* 10^(-8)`fraction of carbon−14 in a piece of charcoal remains after 14.0 years.

b)
`0.000* 10^(-3)`fraction of carbon−14 in a piece of charcoal remains after
`1.900* 10^4 years`

c)
`0.0000* 10^(-4)`fraction of carbon−14 in a piece of charcoal remains after
`1.0000* 10^5 years`.

Step-by-step explanation:

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

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

Taking log both sides:

`\log [A]=t\log[(1)/(2)]-\log [t_{(1)/(2)}]`

[A] = fraction at given time t

`t_{(1)/(2)}` = half life of the carbon−14 =5,730 years

a)When , t = 14 years

`\log [A]=t\log[(1)/(2)]-\log [t_{(1)/(2)}]`

`\log [A]= 14 years* (-3010)-\log [5,730 years]`

`[A]=1.065* 10^(-8)`

b)When , t =
`1.900* 10^4 years`

`\log [A]=t\log[(1)/(2)]-\log [t_{(1)/(2)}]`

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

`[A]=0.000* 10^(-3) [/tex`

c)When , t =
`1.0000* 10^5 years`

`\log [A]=t\log[(1)/(2)]-\log [t_{(1)/(2)}]`

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

`[A]=0.0000* 10^(-4)`