Radioactive Decay — Question 10

Question 10

Detector dead time can make the recorded rate underestimate activity

A nonparalyzable counter becomes insensitive for fixed time tau after each registered event. Events during that interval do not extend it. The observed rate is m and the true arrival rate is R. Treat the mean live fraction as $1-m\tau$.

1. Derive the rate correction and its domain.

2. For $m=5000\,\mathrm{s^{-1}}$ and $\tau=20.0\,\mu\mathrm s$, find R and the fraction of arrivals missed.

Question 10 - Solution

Detector dead time can make the recorded rate underestimate activity

(a) Method and calculation

$$m=R(1-m\tau)\Rightarrow\boxed{R=m/(1-m\tau)}.$$$$\boxed{m\tau<1,\quad m=R/(1+R\tau)<1/\tau}.$$

(b) Interpretation and further calculation

$$\boxed{R=5000/(1-.100)=5555.6\,\mathrm{s^{-1}}}.$$$$\boxed{(R-m)/R=1-m/R=m\tau=.100=10.0\%}.$$

Physical check

This correction belongs to the specified nonparalyzable model. A paralyzable counter has different behavior because unrecorded arrivals can prolong the dead interval.