NUCLEAR DECAY

The activity equation A = dN/dt = -λN is a DIFFERENTIAL EQUATION, as such, it can be solved by calculus though this is not entirely necessary.

We can also solve it through the "half life" version of the decay.

Suppose we consider at the start, N₀ nuclei which are decaying with a half life of T1/2, then;

Number remaining N	time in half lives
N0 = N0 2-0	0 x T1/2
N0 / 2 = N0 2-1	1 x T1/2
N0 / 4 = N0 2-2	2 x T1/2
N0 / 8 = N0 2-3	3 x T1/2
N0 / 2-t/T1/2	t

ie N = N₀ / 2^-t/T1/2

This version describes the decay in terms of numbers of nuclei. But the mass, activity and count rate are all directly proportional to the number of nuclei so we can also write

Fraction remaining

N / N₀ = m / m₀ = C / C₀ = 2^-t/T1/2

We can use log₁₀ on both sides to solve for time if necessary

log₁₀(N / N₀) = -t/T1/2 log₁₀2

If instead we use the "exponential number, e" as a base for logarithms. These logarithms are called "Natural Logs" and "ln" as the symbol. ( We select this strange number called "e" = 2.7172..... , because it arises naturally out of calculus and is OFTEN used in physical expressions.)

loge ( N / N₀ ) = ln ( N / N₀ ) = ln ( 2^-t/T1/2 ) = -( t/T1/2 ) ln 2 = - ( 0.693 / T1/2 ) t

by the definition of a logarithm

N / N0 = e-( 0.693 / T1/2 ) t = m / m0 = C / C0 = A / A0 = 2-t/T1/2 = fraction remaining

If we now bung N in the activity equation A = - dN/dt = λN and differentiate, lo and behold

λ = 0.693 / T1/2

d [ N₀e^{-( 0.693 / T1/2 ) t} ] / dt = N₀ [ de^{-( 0.693 / T1/2 ) t} /dt ] = - ( 0.693 / T1/2 ). N_{0 .} e^{-( 0.693 / T1/2 ) t} = -( 0.693 / T1/2 ) N = -λ N

pheewwee!

All this seems very nasty but problems based on them are dead simple! All you really need is to get used to the "log_e" and "e^x" buttons on your calculator.

Eg
There are 15.3 disintegrations per minute in a 1 gram sample of carbon from C-14 decay. A fossilized tree 1 gram sample gives 1.9 disintegrations per minute. If T1/2 of C-14 is 5700 years approximately, what age is the fossilized wood?

Solution; The fraction remaining of C-14 in the fossilized wood is C / C₀ = 1.9/15.3 = 0.124

So 0.124 = e^{- 0.693 / T1/2 t} = e^{- 0.693 / 5700 t}

thus log_e0.124 = -0.693 / 5700 t = -2.086 using your "log_e " or " ln" button on your calculator.

As the half life is in years, and exponents ( indices ) have no dimensions, the time t will come out in years.

t = 2.086 x 5700 / 0.693 = 17200 years

Clearly this can be done in log₁₀ as well but not as simply!