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 = $\frac{ln2}{t_{1/2}}$

Here, ${t_{1/2}$ is half life of phosphorus 32 = 14.3 days

λ = $\frac{ln2}{14.3}$

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

m=m₀e⁻(λt)

= $4*e^{-\frac{ln2}{14.3}71.5}$

= $4*e^{-ln2*5}$

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

=$4*e^{-ln32}$

=$\frac{4}{32}$

= 0.125 mg

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