In 4 years 20 of a radio active element decays Find its half

In 4 years, 20% of a radio active element decays. Find its half-life.

Solution

A=Pe^kt

A=remaining amount after t years , P=initial amount , t is time in years , k is decay constant

4 years, 20% of a radio active element decays

=>t =4, A=(100%-20%)P=0.8P

0.8P=Pe^k4

e^k4=0.8

4k=ln0.8

k=(1/4)ln0.8

so A=Pe^{((1/4)ln0.8)t}

at half life , 50% of a radio active element decays=>A=(100%-50%)P=0.5P

0.5P=Pe^{((1/4)ln0.8)t}

e^{((1/4)ln0.8)t}=0.5

((1/4)ln0.8)t=ln0.5

t=(ln0.5)_{/((1/4)ln0.8)}

t=12.425 years

half-life=12.425 years