Let us first find out the radioactive constant of Phosphorus 32.
Radioactive constant, λ = [tex] \frac{ln2}{t_{1/2}} [/tex]
Here, [tex] {t_{1/2} [/tex] is half life of phosphorus 32 = 14.3 days
λ = [tex] \frac{ln2}{14.3} [/tex]
The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,
m=m₀e⁻(λt)
= [tex] 4*e^{-\frac{ln2}{14.3}71.5} [/tex]
= [tex] 4*e^{-ln2*5} [/tex]
=[tex] 4*e^{-ln(2)^{5}} [/tex]
=[tex] 4*e^{-ln32} [/tex]
=[tex] \frac{4}{32} [/tex]
= 0.125 mg
The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.
