Uncertainty on half life question?
Naturally occurring samarium includes 15.1% of the radioactive isotope 147Sm,
which decays by α-emission. One gram of natural Sm gives 89 ± 5 α decays per
second. Calculate the half-life of the isotope 147Sm, and give its uncertainty.
- Steve4PhysicsLv 711 months agoFavorite Answer
Work through each step carefully and you’ll see how it’s done.
1g of natural Sm contains 15.1% of 147Sm = 1 x 15.1/100 = 0.151g of 147Sm.
Molar mass of 147Sm = 147g. Divide by Avogadro’s number to get mass of 1 atom:
m = 147/(6.022x10^23) = 2.441x10^-22g
(Alternatively, mass of a 147Sm atom is 147u = 147 x 1.6605x10^-27 = 2.441x10^-25 kg = 2.441x10^-22g.)
0.151g of 147Sm contains N atoms of 147Sm where:
N = 0.151/(2.441x10^-22) = 6.186x10^20 atoms
Activity, A = λN where λ is the decay constant. Since λ = ln(2)/T_half
A = Nln(2)/T_half
T_half = Nln(2)/A (equation 1)
Using A = 89 dps gives:
T_half = 6.186x10^20 x ln(2)/ 89 = 4.82*10^18s
You can deal with the uncertainty in 2 ways which give slightly different results.
A = 89±5 dps which is a fractional uncertainty of 5/89 = 0.056 = 5.6%
Since we have simply divided by A, the fractional uncertainty of the result is also 5.6%.
5.6% of 4.82*10^18s = 2.70x10^17 = 0.27*10^18s
T_half = (4.82±0.27)x10^18s
It makes sense to round the uncertainty to 1 sig. figure:
T_half = (4.8±0.3)x10^18s
(The answers are large but 147Sm has a very long half-life.)
I expect that’s the method you are meant use, but for information...
Use equation 1 with A = 89-5 = 84dps to get an upper limit for T_half.
T_half_max = 6.186x10^20 x ln(2)/ 84 = 5.1*10^18s
Use equation 1 with A = 89+5 = 94dps to get a lower limit for T_half.
T_half_min= 6.186x10^20 x ln(2)/ 94 = 4.6*10^18s
Check my working/arithmetic of course.
- BillLv 711 months ago
Using Avogadro's number and the given percentage 15.1 percent, calculate how many atoms of the radioactive isotope are in a one gram sample. If 89 atoms decay per second, how long would it take half of the atoms to decay?