Question

The dead time of a counter system is to be determined by taking measurements on two radioactive sources individually and collectively. If the pulse counts over a time interval t are, respectively, N₁, N₂ and N₁₂, what is the value of the dead time?

Answer 1

The problem is based on the double source method for the determination of the dead time of a G.M. counter. Two radioactive sources of comparable strength are chosen. In all four counts are taken. First, the background rate B per second is found out when neither source is present. One of the sources is placed in a suitable position so that a high counting rate N₁ is registered. While this counter is in the same position, the second source is placed by its side to get the counting rate N₁₂. Finally, the first source is removed so that the second counter above gives a count of N₂. If n₁, n₂ and n₁₂ are the true counting rates then we expect 

(n₁ + B) + (n₂ + B) = n₁₂ + B

or 

n₁ + n₂ = n₁₂ − B (1) 

Now, for each particle that is counted, on an average there will be a dead time τ during which particles are not counted. Then if N is the counting rate (number of counts/second) then the time lost will be Nτ , so that the true counting rate 

n = N/(1 − Nτ ) 

Thus n₁ = N₁/(1 − Nτ) etc. 

We can then write (1) as N₁/(1 − N₁τ) + N₂/(1 − N₂τ) = N₁₂/(1 − N₁₂τ) − B/(1 − Bτ) (2) 

Now, in practice N₁ and N₂ will be of the order of 100 per second, N₁₂ is of 200 per second, B ≈ 1 per second and τ ≈ 10⁻⁴ second, so that N₁ τ << 1, etc. We can then expand the denominators binomially and write to a good approximation 

N₁(1 + N₁τ ) + N₂(1 + N₂τ ) = N₁₂(1 + N₁₂τ ) − B(1 + Bτ) 

N₁ + N₂ − N₁₂ + B = τ[N₁₂² − N₁² − N₂² ] ≈ τ [(N₁ + N₂)² − N₁² − N₂²] 

Or τ = (N₁ + N₂ − N₁₂ + B)/2N₁N₂