Question 42.35: A certain radionuclide is being manufactured in a cyclotron ......
A certain radionuclide is being manufactured in a cyclotron at a constant rate R. It is also decaying with disintegration constant λ. Assume that the production process has been going on for a time that is much longer than the half-life of the radionuclide. (a) Show that the number of radioactive nuclei present after such time remains constant and is given by N = R/λ. (b) Now show that this result holds no matter how many radioactive nuclei were present initially. The nuclide is said to be in secular equilibrium with its source; in this state its decay rate is just equal to its production rate.
THINK We modify Eq. 42-11 to take into consideration the rate of production of the radionuclide.
M=m_1 m_2 (34-11)
EXPRESS If N is the number of undecayed nuclei present at time t, then
\frac{d N}{d t}=R-\lambda N
where R is the rate of production by the cyclotron and λ is the disintegration constant. The second term gives the rate of decay. Note the sign difference between R and λN.
ANALYZE (a) Rearrange the equation slightly and integrate:
\int_{N_0}^N \frac{d N}{R-\lambda N}=\int_0^t d t
where N_0 is the number of undecayed nuclei present at time t = 0. This yields
-\frac{1}{\lambda} \ln \frac{R-\lambda N}{R-\lambda N_0}=t \text {. }
We solve for N:
N=\frac{R}{\lambda}+\left(N_0-\frac{R}{\lambda}\right) e^{-\lambda t}
After many half-lives, the exponential is small and the second term can be neglected. Then, N = R/λ.
(b) The result N = R/λ holds regardless of the initial value N_0, because the dependence on N_0 shows up only in the second term, which is exponentially suppressed at large t.
LEARN At times that are long compared to the half-life, the rate of production equals the rate of decay and N is a constant. The nuclide is in secular equilibrium with its source.