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

Activity (A) = 1.5 Ci
The decay constant (λ) of technetium-99m is 2.21 x 10^-6 s^-1. 
The number of disintegrations in time t can be calculated using the following equation:

N = A/ λ * (1 – e^-λt)

Substituting the given values, we get:

N = (1.5 Ci) / (2.21 x 10^-6 s^-1) * (1 – e^-(2.21 x 10^-6 s^-1 * 5.0 s))
 
N = 2.8 x 10^11 disintegrations

Therefore, the number of disintegrations that occur in the technetium-99m sample in 5.0 sec is 2.8 x 10^11. The answer is B.

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