A patient swallows a radiopharmaceutical tagged with phosphorus-32( 32 15 P),a -

Question

A patient swallows a radiopharmaceutical tagged with phosphorus-32(3215P),a - emitter with ahalf-life of 14.3 days. The average kinetic energy of the emittedelectrons is 700 keV. The initial activity of the sample is 1.31MBq. (a) What is the number of electrons emitted in a 8.0 day period?
electrons

(b) What is the total energy deposited in the body duringthe 8.0 days?
J

(c) What is the absorbed dose if the electrons are completelyabsorbed in 92 g oftissue?
rad (a) What is the number of electrons emitted in a 8.0 day period?
electrons

(b) What is the total energy deposited in the body duringthe 8.0 days?
J

(c) What is the absorbed dose if the electrons are completelyabsorbed in 92 g oftissue?
rad

Explanation / Answer

   from the theory we have the equation    N = R /        = Roe- t /        = (T1/2Ro / ln2) (1 - e- t ln2 /T1/2)    so the number of decays occuring during the 9.0day period is    N = No - N          =........ decays (b)    the total energy deposited will be    E = (700 keV / decay) (N) (1.602 x10-16 J / 1 keV)       = ......... J (c)    so the total absorbed dose will be    dose = energy deposited per unit mass / energydeposited per rad            =(E / 0.920 kg) / (10-2 J/kg / rad)            =........ rad

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.