- #1
moenste
- 711
- 12
Homework Statement
When iron is irradiated with neutrons an isotope of iron is formed. This isotope is radioactive with a half-value period (half-life) of 45 days. Give the meanings of the terms printed in italics.
A steel piston ring of mass 16 g was irradiated with neutrons until its activity due to the formation of this isotope was 10 microcurie. Ten days after the irraditation the ring was installed in an engine and after 80 days continuous use the crankcase oil was found to have a total activity of 1.85 * 103 disintegrations per second. Determine the average mass of iron worn off the ring per day assuming that all the metal removed from the ring accumulated in the oil and that one curie is equivalent to 3.7 * 1010 disintegrations per second.
Answer: 4.0 mg per day.
2. The attempt at a solution
First of all I calculated the decay constant λ:
A = A0 e- λ t
1850 = 370 000 e- λ 90 days or 7 776 000 s
λ = 6.8 * 10-7 s-1
I then calculated the number of atoms at t = 0 and t = 90 days:
dN / dt = - λ N
N0 = 370 000 / 6.8 * 10-7 = 5.4 * 1011 atoms.
N90 = 1850 / 6.8 * 10-7 = 2 720 588 235 atoms
Then I found the relative atomic number Ar: N = m NA / Ar → Ar = m NA / N = 16 * 6 * 1023 / 5.4 * 1011 = 1.78 * 1013
Then I found the mass at t = 90: m = Ar N / NA = 1.78 * 1013 * 2 720 588 235 / 6 * 1023 = 0.08 g.
And now find the difference 16 g - 0.08 g = 15.92 g mass worn off in total. 15.92 / 90 = 0.177 g per day worn off.
Not 0.004 g per day, what's wrong?