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

Since the dead-time losses here are small (\(N_{obs}\tau=0.075\) for the gross rate and 0.025 for the background, i.e. 7.5% and 2.5% losses), we can shortcut the division by using the first-order binomial expansion \(\dfrac{1}{1-x}\approx1+x\) for \(x\ll1\), instead of dividing directly:

\[ N_{true}\approx N_{obs}(1+N_{obs}\tau) \]

Gross rate: \(N_{true,\,gross}\approx300(1+0.075)=300\times1.075=322.5\ \text{cps}\)
Background: \(N_{true,\,bg}\approx100(1+0.025)=100\times1.025=102.5\ \text{cps}\)

Net rate: \(322.5-102.5=220.0\ \text{cps}\)

This quick approximate route lands at exactly 220.0 cps, while the exact division above gave 221.76 cps — both values sit comfortably inside the official 220–223 cps window, confirming that the dead-time correction must be applied to the background as well as to the gross signal before the subtraction, not to the gross count alone.