What is the half-life of thallium-209 if 5.63 minutes are required for the activity of a sample of thallium-209 to fall to 17.0 percent of its...

To solve, apply the formula:

`A=A_o * 2^(-t/T_(1/2))`

where

A is the remaining amount

Ao is the original amount

t is the age

T_1/2 is the half-life

Plugging in A=0.17Ao and t=5.63, the formula becomes:

`0.17A_o = A_o *2^(-5.63/T_(1/2))`

`0.17=2^(-5.63/T_(1/2))`

Then, take the LN of both sides.

`ln(0.17) = ln (2^(-5.63/T_(1/2)))`

`ln (0.17)=-5.63/T_(1/2) *ln(2)`

And isolate T_1/2.

`T_(1/2)*ln (0.17) = -5.63 * ln(2)`

`T_(1/2) = (-5.63*ln(2))/(ln(0.17))`

`T_(1/2)=2.20`

Therefore, the half-life of Thallium-209 is 2.20 minutes.

Comments

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