Find the half-life (in hours) of a radioactive substance that is reduced by 40 percent in 35 hours

Answer 1

So if it is reduced by 40% in 35 hours then it has 60% remaining after 35 hours. Using our basic formula

A=A₀e^kt

If our starting amount (A₀) is 100 then our amount remaining (A) is 60, if our starting amount (A₀) is 1 then our amount remaining (A) is 0.6. We can also just leave A₀ and use A=0.6(A₀). Whatever makes the most sense to you.

Using A=0.6(A₀), t=35

0.6(A₀)=A₀e^k(35)

Divide both sides by A₀ (This step whatever choice you made for your starting amount will cancel out and we will get the same result.

0.6=e^k(35)

Take the ln of both sides

ln(0.6)=35k

divide by 35

k=ln(0.6)/35

plug this into equation

A=A₀e^{(ln(0.6)/35)t}

Now that we have the expression we can solve for t when A=0.5A₀ (Or A=0.5 A₀=1 Or A=50 A₀=100)

0.5A₀=A₀e^{(ln(0.6)/35)t}

Follow same steps as above but this time we are solving for t

0.5=e^{(ln(0.6)/35)t}

Take ln of each

ln(0.5)=(ln(0.6)/35)t

Multiplty by 35, divide by ln(0.6)

35ln(0.5)/ln(0.6)=t

t≈47.5hrs

You could have found k and rounded to a decimal and used that instead of tracking through ln(0.6)/35 if you prefer.