Question

A radioactive isotope of phosphorus, ³²P, is used in internal radiotherapy to treat patients with cancers and other diseases of the bone marrow. ³²P has a half-life of 14.28 days and a atomic mass of ³²P is 31.973 u. Via β decay, it emits an electron with an energy of 705 KeV and a RBE of 15.5 .

As part of a treatment program, a patient ingests a radioactive pharmaceutical containing ³²P which has an initial activity of 1.33 MBq.

1) How many electrons are emitted over a period of 7 days?

N =

2) What is the total amount of energy absorbed by the patients body during this period?

E_absorbed =

3) If the radiation is absorbed by 123 g of tissue, what is the patient's equivient absorbed dosage?

Dose =

Answer 1

リ derey 0.0485294 / day Tyz Anzano (an contant) Activity, 0.693 0.693 14.28 Ao 1.33x10 (0.0485294 94/24x60x60) Noz 2.3679x 10² No ett -0.048529487 2.3679x10²x (Ideya) 1.6859 x 10" N2 12 N = No. 6.9 elections emitted No-N Ne z 2.3679 x 1012 1.6859x10²=0.682x10. Nea 6.82x10 elections 2) E abrobed a Nex eletion a 6.82x10x7.05kev enegy of Eabrorbed 6.82x10 x 7.05x13x 1.602xio → Fabrabed a 7.7026x104 I Eabarbed 6.2623 x 10 Man 7.7026x104 0.123 Gray Dore = 6.2623 x 103 x 15.5 = 0.097066 Sievert.