Question

Phosphorus 32 has a half life of 14.3 days how much of a 4mg sample of phosphorus - 32 will remain after 71.5 days

Answer 1

Let us first find out the radioactive constant of Phosphorus 32.

Radioactive constant, λ =
`(ln2)/(t_(1/2))`

Here,
`{t_(1/2)` is half life of phosphorus 32 = 14.3 days

λ =
`(ln2)/(14.3)`

The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,

m=m₀e⁻(λt)

=
`4*e^{-(ln2)/(14.3)71.5}`

=
`4*e^(-ln2*5)`

=
`4*e^{-ln(2)^(5)}`

=
`4*e^(-ln32)`

=
`(4)/(32)`

= 0.125 mg

The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.