Question

A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in days.

Answer 1

The substance half-life is of 4.98 days.

Explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

`A(t) = A(0)e^(-kt)`

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

`A(t) = A(0)e^(-0.1392t)`

Find the substance's half life, in days.

This is t for which
`A(t) = 0.5A(0)`. So

`A(t) = A(0)e^(-0.1392t)`

`0.5A(0) = A(0)e^(-0.1392t)`

`e^(-0.1392t) = 0.5`

`\ln{e^(-0.1392t)} = ln(0.5)`

`-0.1392t = ln(0.5)`

`t = -(ln(0.5))/(0.1392)`

`t = 4.98`

The substance half-life is of 4.98 days.