Question

A scintillometer records 300 counts per second (cps) in a radiometric survey. If the background radiation and dead-time of the instrument are 100 cps and 250 \(\mu s\), respectively, then the true net count rate is _______ cps (rounded off to two decimals).

Hint:

Apply the non-paralysable dead-time correction \(N_{true}=N_{obs}/(1-N_{obs}\tau)\) to BOTH the gross count rate and the background count rate before subtracting them.

Answer 1

Correct Answer: 220

A scintillometer always under-counts at high count rates because the detector needs a finite dead-time \(\tau\) to recover after each pulse before it can register the next one. For a non-paralysable (Type I) counter, the observed (recorded) count rate \(N_{obs}\) and the true count rate \(N_{true}\) are related by:

\[ N_{true}=\dfrac{N_{obs}}{1-N_{obs}\tau} \]

Step 1: Correct the gross (signal + background) rate.
Here \(N_{obs}=300\) cps and \(\tau=250\ \mu s=2.5\times10^{-4}\ s\).
\(N_{obs}\tau=300\times2.5\times10^{-4}=0.075\)
\(N_{true,\,gross}=\dfrac{300}{1-0.075}=\dfrac{300}{0.925}=324.32\ \text{cps}\)

Step 2: Correct the background rate the same way.
The quoted background of 100 cps is itself a value recorded on the same dead-time-limited instrument, so it must also be corrected before being subtracted:
\(100\times2.5\times10^{-4}=0.025\)
\(N_{true,\,bg}=\dfrac{100}{1-0.025}=\dfrac{100}{0.975}=102.56\ \text{cps}\)

Step 3: Net (source-only) true count rate.
\(N_{net}=N_{true,\,gross}-N_{true,\,bg}=324.32-102.56=221.76\ \text{cps}\)

\[ \boxed{N_{net}\approx221.76\ \text{cps}} \]

This lies inside the accepted 220–223 cps band.