Introduction:
In this problem, we are given information about the decay of a radioactive isotope, 90Th232, and the count rate measured by an alpha detector in an experiment. We are required to determine the efficiency of the alpha detector.

Given:
- Mass of 90Th232 = 1 gram
- Average count rate measured by the alpha detector = 3000 counts/s
- Half-life of 90Th232 = 4.4 × 10^17 s

Efficiency of the Alpha Detector:
The efficiency of a detector is defined as the ratio of the number of counts detected by the detector to the total number of particles emitted by the source. In this case, the alpha detector measures the number of alpha particles emitted by the decay of 90Th232.

The decay of 90Th232 follows the decay law, where the number of radioactive nuclei N at a given time t is given by the equation:

N(t) = N0 * e^(-λt)

where N0 is the initial number of radioactive nuclei, λ is the decay constant, and t is the time.

Calculating the Efficiency:
To calculate the efficiency of the alpha detector, we need to determine the total number of alpha particles emitted by the decay of 90Th232 and the number of alpha particles detected by the alpha detector.

Step 1: Determining the Total Number of Alpha Particles Emitted:
The total number of radioactive nuclei N0 can be calculated using the given mass of 90Th232 and Avogadro's number (6.022 × 10^23 mol^-1).

Number of moles of 90Th232 = mass / molar mass
Number of moles = 1 g / (232 g/mol) = 0.00431 mol

Total number of 90Th232 nuclei = number of moles * Avogadro's number
N0 = 0.00431 mol * (6.022 × 10^23 mol^-1) = 2.6 × 10^21 nuclei

The number of alpha particles emitted by the decay of 90Th232 is equal to the total number of radioactive nuclei N0.

Number of alpha particles emitted = N0 = 2.6 × 10^21

Step 2: Determining the Number of Alpha Particles Detected:
The number of alpha particles detected by the alpha detector is given as 3000 counts/s. Since the count rate is given as the average count rate integrated over the entire volume, we can assume that all the emitted alpha particles are detected.

Number of alpha particles detected = 3000 counts/s

Step 3: Calculating the Efficiency:
Efficiency = (Number of alpha particles detected / Number of alpha particles emitted) * 100

Efficiency = (3000 / 2.6 × 10^21) * 100

Efficiency ≈ 1.15 × 10^-19 %

Conclusion:
The efficiency of the alpha detector in this experiment is approximately 1.15 × 10^-19 %. This means that only a very small fraction of the alpha particles emitted by the decay of 90Th232 are detected by the alpha detector.