A sample of technetium-99m has an activity of 1.5 Ci. How many disintegrations occur in the technetium-99m sample in 5.0 sec?

Question

Quizplus · Accepted Answer

The activity of the sample is given by the number of disintegrations per second. This means that we can find the number of disintegrations in 5.0 seconds by multiplying the activity (in Ci) by the time (in seconds).

1 Ci = 3.7 × 10^10 disintegrations per second

Therefore, the number of disintegrations in 5.0 seconds is:

1.5 Ci × (5.0 seconds) × (3.7 × 10^10 disintegrations per second) = 2.8 × 10^11 disintegrations

A Sample of Technetium-99m Has an Activity of 1