Question

A 50.0 mg sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 19.7 days, the amount of radioactive substance had decreased to 3.13 mg. What is the half-life of the unknown radioactive substance?

Answer 1

Answer: 4.928 days

Explanation:

This problem can be solved using the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=3.13mg$ is the final amount of the material

$A_{o}=50mg$ is the initial amount of the material

$t=19.7days$ is the time elapsed

$h$ is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find $h$ from (1):

$3.13mg=(50mg)2^{\frac{-19.7days}{h}}$ (2)

$\frac{3.13mg}{50mg}=2^{\frac{-19.7days}{h}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{3.13mg}{50mg})=ln(2^{\frac{-19.7days}{h}})$ (4)

$-2.77=-\frac{19.7days}{h}ln(2)$ (5)

Clearing $h$:

$h=\frac{-19.7days}{-2.77}(0.693)$ (6)

Finally:

$h=4.928days$