Finding the Half-Life of a Radioactive Element

[Conrad Peterson]

Homework Statement

In 2 years, 20% of a radioactive element decays. Find its half-life rounded to 2 decimal places.

Homework Equations

A(t)=a*e^kt

The Attempt at a Solution

The only thing I've been able to figure out is k=LN(0.2)/2

 

[Charles Link]

The time of the half life is defined as the $t$ such that $\frac{A(t)}{a}=\frac{1}{2}$. But, please show your work. I think you computed $k$ correctly, but it is easier for the homework helper if you show your work. Otherwise, it is hard to tell if you are simply using formulas, or if you know how to do the necessary computations. $\\$ Editing: And if 20% decays, that means that $\frac{A(2 \, years)}{a}=.8$ . Upon computing it myself, I see your answer for $k$ is incorrect. And also, with the computation of $k$, it is necessary to keep the units, such as "years" in the denominator.

 

[symbolipoint]

THIS IS WRONG AND CORRECTION TO APPEAR IN POST #6...
Your are on the right track. For a model y=a*e^(-kt), you are taking first, y=0.2, a=1, and x=2.
e^(-2k)=0.2
then take natural log of both sides, and ...
k=-ln(0.2)/2 or k=0.8047.

The more specific model becomes A(t)=a*e^(-0.8047t).

Find the half-life?
This means, 1*e^(-0.8047t)=1/2;
Solve for t.
-
SEE POST #6 FOR CORRECTION

 

[Charles Link]

Your are on the right track. For a model y=a*e^(-kt), you are taking first, y=0.2, a=1, and x=2.
e^(-2k)=0.2
then take natural log of both sides, and ...
k=-ln(0.2)/2 or k=0.8047.

The more specific model becomes A(t)=a*e^(-0.8047t).

Find the half-life?
This means, 1*e^(-0.8047t)=1/2;
Solve for t.

If 20% has decayed, that means 80% remains. The radioactive decay formula gives the part that remains, so that $e^{-2k}=.8$ where the 2 in the exponent is 2 years.

 

[symbolipoint]

If 20% has decayed, that means 80% remains. The radioactive decay formula gives the part that remains, so that $e^{-2k}=.8$ where the 2 in the exponent is 2 years.

I will take a careful look. I have recently been making very frequent arithmetic mistakes.

 

[symbolipoint]

This is unfinished correction for post #3.
For 2 years time pass, 20% decayed meaning 80% still present.
y=a*e^(-kt)
0.8=1*e^(-2k)
Take natural log both sides,...
k=-ln(0.8)/2
Value for k is 0.1116.

Model to use, A(t)=a*e^(-0.1116t)

Next, finding the half-life.
Say, a=1, A(t)=1/2.
Now, e^(-0.1116t)=1/2.
Take natural log both sides and solve for t.
That will be the half-life.