You have just isolated a new radioactive element. If you can determine its half-life, you will win the Nobel Prize in physic. You purify a sample of 2 grams.One of you colleges steals half of it, and 3 days later you find that 0.1 grams of the radioactive material is still left. What is the half life?
Heres what I did but I don't think its right I am very confused on how to set this up to solve. Please if this is not right help me set this up properly. Thank you.
R(t)=Ae^-kt
0.1=2e^-k3
0.1/2=e^-k3
ln 0.1/2=-k3
ln0.1/2/3=-k
-k=0.998577425/-1
k=0.998577425
That's a lot of work. Just think about it a bit. You start with 1 gram. 3 days later the amount left is 0.1 gram
If the half life is k days, then
(1/2)^3k = 1/10
3k = log(.1)/log(.5) = 3.32
So, k = 1.11
R(t) = (1/2)^(1.11t)
But, since we like things base e for our calculators, that is
e^((-ln2)*1.11)t) = e^-.77t
check: e^(-.77*3) = 0.1
You started off wrong, because you forgot that half of it was stolen. Your work is right, but you should have A=1, not A=2.
To determine the half-life of the radioactive element, you can follow these steps:
1. Start with the radioactive decay equation:
R(t) = A * e^(-kt)
where R(t) represents the remaining amount of the radioactive material at time t, A is the initial amount, e is the base of the natural logarithm, k is the decay constant, and t is the time elapsed.
2. You know that the initial amount is 2 grams, and after 3 days, 0.1 grams of the material is still left. So, you can set up the equation:
0.1 = 2 * e^(-k * 3)
3. Divide both sides of the equation by 2 to isolate the exponential term:
0.1/2 = e^(-k * 3)
4. Take the natural logarithm (ln) of both sides to eliminate the exponential term:
ln(0.1/2) = -k * 3
5. Solve for -k by dividing the left side of the equation by -3:
ln(0.1/2) / -3 = k
6. Plug in the values into a calculator (or use software) to evaluate the expression:
k ≈ 0.998577425 (rounded to the appropriate number of decimal places)
Therefore, the half-life of the radioactive element is approximately equal to 0.998577425.