Find the​ half-life of a radioactive​ element, which decays according to the function ​A(t)​, where t is the time in years.

Question

Find the​ half-life of a radioactive​ element, which decays according to the function

​A(t)=A0e−0.0189t​,

where t is the time in years.

Answer 1

As a math and physics teacher I believe we must have one math approach for physics phenomenon. I mean the same math description.

It is given to us that at any time t the number of decayed particle A(t) = A₀e^{- 0.0189t} .

Here

A₀ - the initial amount of particles,

λ = 0.0189 - decay constant

By definition half-life T_1/2 is a time of radioactive decay of half of the particles. It means after time T_1/2 we will have A₀/2 particles left.

Hence

A₀/2 = A₀e^-λT_1/2.

After simplification we see that

e λT_1/2 = 2

Taking natural logarithm from both sides, we get:

λT_1/2 = ln 2

or

T_1/2 = 0.693 / λ

Hope you understand that 0.693 is the natural log of 2.

The formula that we deduced could be found in any physics textbook.

Putting the value for λ = 0.0189, we obtain

T_1/2 = 0.693 / 0/0189 = 36.667 years.

We do not need greater precision because in ln 2 we usually use only 3 decimal places.

Answer 2

When 1 half life goes by, then half of the original material (A₀) will be gone. That means there is 1/2A₀ left. Plug in 1/2A₀ for A(t) and solve for "t":

1/2A₀ = A₀e^-0.0189t

Now divide both sides by A₀ to get:

1/2A₀/A₀ = e^-0.0189t

1/2 = e^-0.0189t

Now take the natural log of both sides:

ln(1/2) = ln(e^-0.0189t)

One of the rules of logarithms is that you can move the exponent out front to be a multiplier:

ln(1/2) = (-0.0189t)ln(e) and then since ln(e) = 1:

ln(1/2) = -0.0189t

ln(1/2)/(-0.0189) = t

Plugging this into a calculator gives t = 36.674454

You can round to whatever suites your fancy.