If a Mo99/Tc-99m generator has an elution efficiency of 96% and contains 11.7 MBq of Tc99m, how much activity will be removed during elution?

To find the amount of activity that will be removed during elution from a Mo99/Tc-99m generator, you multiply the total activity of Tc-99m by the elution efficiency. The generator has 11.7 MBq of Tc-99m and an elution efficiency of 96%.

The calculation for the activity removed during elution would be:

Activity removed = Total activity of Tc-99m × Elution efficiency

= 11.7 MBq × 0.96

= 11.232 MBq.

When rounding to one decimal place, this results in approximately 11.2 MBq. Therefore, option B is the correct answer as it correlates with the calculated activity removed during elution.

This demonstrates the importance of understanding both the total available activity and the concept of elution efficiency when dealing with radiopharmaceuticals in practice, as these calculations are critical for ensuring accurate dosing and effective use of the radiopharmaceuticals in medical applications.