Mathematics: Cerium-144 is a radioactive isotope with a half-life of 285 days. how long would it take for an initial sample of 15 grams of cerium-144 to decay until only 5 gram remains? a. about 452 days b. about 855 days c. about 570 days d. about 407 days

The answer will be X = 24, -24...

Cerium-144 is a radioactive isotope with a half-life

Ответ:

mazolethrin6876

14.03.2022

Mathematics

To solve this we are going to use the formula: $A=A_{0}( \frac{1}{2} )^{ \frac{t}{h}$
where

is the final amount aster a time

$A_{0}$ is the initial amount

is the time

is the half-life

We know for our problem that A=5

, $A_{0}=15$ , and n=285

, so lets replace those values in our formula:
$A=A_{0}( \frac{1}{2} )^{ \frac{t}{h}$
$5=15( \frac{1}{2} )^{ \frac{t}{285}$

Since

is the exponent, we are going to use logarithms to find its value:
$5=15( \frac{1}{2} )^{ \frac{t}{285}$
$( \frac{1}{2} )^{ \frac{t}{285}}= \frac{5}{15}$
$( \frac{1}{2} )^{ \frac{t}{285}}= \frac{1}{3}$
$ln( \frac{1}{2} )^{ \frac{t}{285}}= ln(\frac{1}{3})$
$\frac{t}{285} ln( \frac{1}{2})=ln( \frac{1}{3} )$
$\frac{t}{285}= \frac{ln( \frac{1}{3}) }{ln( \frac{1}{2} )}$
$t= \frac{285ln( \frac{1}{3}) }{ln( \frac{1}{2}) }$
t=451.71

We can conclude that the correct answer is: a. about 452 days.

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Cerium-144 is a radioactive isotope with a half-life of 285 days. how long would it take for an initial sample of 15 grams of cerium-144 to decay until only 5 gram remains? a. about 452 days b. about 855 days c. about 570 days d. about 407 days

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