A tracer has half-life 6.0 h and initial activity 48 MBq. A 20 min scan begins 12 h later. (a) Find activity at scan start and end. (b) Estimate decays using mean activity. (c) For one 140 keV photon per decay, estimate emitted gamma energy using 1 eV = 1.60 × 10⁻¹⁹ J. (d) Explain suitable gamma and half-life properties. (e) Explain attenuation, collimation and optimisation. (f) Distinguish activity from absorbed dose.

Question

Accepted Answer

Start 12.0 MBq; end 11.55 MBq; about 1.41 × 10¹⁰ decays; emitted gamma energy about 3.16 × 10⁻⁴ J. Twelve hours is two half-lives, so start activity = 48/4 = 12.0 MBq. Over 1/3 h, decay factor 2^(−1/18) = 0.9622, giving 11.55 MBq. Mean activity ≈11.775 MBq = 1.1775 × 10⁷ s⁻¹. Over 1200 s, decays ≈1.41 × 10¹⁰. Photon energy = 140000 × 1.60 × 10⁻¹⁹ = 2.24 × 10⁻¹⁴ J; total emitted energy ≈3.16 × 10⁻⁴ J. Gamma penetrates tissue to an external detector; an hours-long half-life supports preparation and imaging but then falls, reducing prolonged exposure. Attenuation lowers counts; collimation maps direction but rejects photons. Optimise administered activity and scan time while staff minimise time, maximise distance and shield. Activity is decays per second. Absorbed dose is deposited energy per unit mass and also depends on energy, geometry and absorption; not every emitted photon is absorbed.

Nuclear tracer dose and gamma imaging

What a complete response should do

How did this one feel?