The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many grams were in the original sample?

Jul 3, 2016

${m}_{i} = 400 \setminus g r a m s$

Explanation:

The following formula relates the mass remaining of the radioiosotope to the original mass:

${m}_{i} = {m}_{r} \times {2}^{n}$

Where:

${m}_{i} : \text{ is the initial mass of the radioisotope}$
${m}_{r} : \text{ is the mass remaining of the radioisotope after n periods}$
$n : \text{ is the number of periods}$
$n = \left(\text{time")/ ("half life}\right)$

First, find the number of periods.

$n = \left(15.8 \text{ years")/(5.26 " years}\right)$
$n = 3$
Then plug in the values in the original formula

${m}_{i} = {m}_{r} \times {2}^{n}$
${m}_{i} = 50 \times {2}^{3}$
${m}_{i} = 400 \setminus g r a m s$

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A quick approach

Knowing that the time represents three periods

$50 \to 100 \to 200 \to 400 \setminus g r a m s$