PLEASE SHOW ACTUAL NUMBERS AND FORMULAS WHENANSWERING PLEASE A patient swallows

Question

PLEASE SHOW ACTUAL NUMBERS AND FORMULAS WHENANSWERING PLEASE
A patient swallows a radiopharmaceutical tagged with phosphorus-32(3215P), a -emitter with a half-life of 14.3 days. The average kinetic energyof the emitted electrons is 700 keV. The initial activity of thesample is 1.31 MBq. (a) What is the number of electrons emitted ina 7.5 day period?
electrons

(b) What is the total energy deposited in the body during the7.5 days?
J

(c) What is the absorbed dose if the electrons are completelyabsorbed in 104 g of tissue?
rad PLEASE SHOW ACTUAL NUMBERS AND FORMULAS WHENANSWERING PLEASE
A patient swallows a radiopharmaceutical tagged with phosphorus-32(3215P), a -emitter with a half-life of 14.3 days. The average kinetic energyof the emitted electrons is 700 keV. The initial activity of thesample is 1.31 MBq. (a) What is the number of electrons emitted ina 7.5 day period?
electrons

(b) What is the total energy deposited in the body during the7.5 days?
J

(c) What is the absorbed dose if the electrons are completelyabsorbed in 104 g of tissue?
rad

Explanation / Answer

(a)    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 duringthe 7.5 day 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 / 0.104 kg) / (10-2 J/kg / rad)            =........ rad

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