Using the decay equation:
N(t)=N0 * 2^(-t/t_0.5)
where N0 = number of atoms initially, N(t) = number of atoms
remaining after time t, and t_0.5 is the half life, we see that the
proportion of atoms remaining after time t, R(t) is given by:
R(t) = N(t)/N0 = 2^(-t/t_0.5)
and the proportion consumed C(t) is thus:
C(1) = 1 - R(t) = 1 - 2^(-t/t_0.5)
If t=1 year, and t_0.5 = 76000 years, then
C(1 year) = 1 - 2^((1 y)/(76000 y) = 9.12 x 10^-6
Given the logarithmic curve is flat up front, and 1 year = 5.26x10^5
min, we have the approximation:
C(1 min) = C(1 year)/(5.26x10^5) = 9.12x10^-6 / 5.26x10^5
C(1 min) = 1.734 x10^-11
A mole of material with the given decay rate will produce radiation
at a dpm/mol rate Cm where:
Cm = Na * 1.734 x10^-11 / min = (6.022x10^23 / mol)*(1.734 x10^-11)
Cm = 1.044x10^13 dpm/mol = 1.74x10^11 Bq/mol
If Rossi's experiment produced 3 grams of copper, it should produce
about 7/3 as much 59Ni, or about 7 grams of 59Ni.
Using 59 g/mol, the 7 grams is 7/59 = 0.1186 mol, so represents
(1.74x10^11 Bq/mol)(0.1186 mol) = 2.03x10^10 Bq,
Using a counting efficiency even as low as 0.01, there should be a
count rate of about 2x10^8 / s.
That would be unmistakable, to say the least!
My intuition on this earlier was wrong because I am used to very
small proportions of isotopes of an element being radioactive. This
is pure stuff. It pays to do the math!
So, you are right Jones, if 59Ni is present then it should be evident!
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
