Question #03416

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Question #03416

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Answer 1 · 2017-04-14T16:59:01

11460 years.

Explanation:

$N/(N_0) = (1/2)^n$

Where, $N_0$ = Original amount of radioactive nuclei present.

$N$ = Nuclei present at that instant.

$n = t/(T)$

$T$ = Half life of Carbon-14
Half life of C-14 is 5730 years.

$t$ = Time in which the particular number of nuclei has disintegrated.

$:. N/(N_0) = (25/100)$

Which is,

$(1/2)^n = (1/4)$

$(1/2)^n = (1/2)^2$

$t/T = 2$

$t = T × 2$

$t = 5730 × 2$

$t = 11460$ years

Answer 2 · 2017-04-19T04:52:38

$11460$ years old

Explanation:

At time $t=0$ , we have $100%$ of the radioactive $""^14"C"$ isotope.

The half-life of $""^14"C"$ is $5730$ years. This means that at $t=5730$ years, only $50%$ of the original $""^14"C"$ remains.

Taking half of $50%$ , we see that the next half life will occur at $t=2xx5730=11460$ years. At this point in time, only $25%$ of the $""^14"C"$ will remain.

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Question #03416

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