Question

A 95 kg patient swallows a 34 ?Ci beta emitter whose half-life is 5.0 days and whose RBE is 1.6. The beta particles are emitted with an average energy of 0.35 MeV, 90% of which is absorbed by the body.

You are a health care worker needing to find the patient's dose equivalent after one week. These series of steps will help you find that dose equivalent. In all questions, assume the radioactive nuclei are distributed throughout the patient's body and are not being excreted.

Part A

How many radioactive nuclei did the patient swallow?

N0= 7.84x10^11

Part B

How many radioactive nuclei are left in the patient after one week?

N=2.97x10^11

Part C

How many radioactive decays occured in the patient during the first week?

ND= 4.87x10^11

Part D

How much energy (in Joules) was deposited into the patient during the week?

deltaE = 2.45x1-^-2

Part E

What dose equivalent (in mSv) does the patient receive in the first week? (To determine whether you should be concerned about your patient, remember that natural background exposure is about 3 mSv.)

  

Answer 1

A)

dN/dt = - $\lambda$ N

dN/dt = rate of decay

dN/dt = 34 x 10^-6 Ci

dN/dt = 34 x 10^-6 x 3.7 x 10^10 Bq

dN/dt = 125.8 x 10^4 Bq

N = dN/dt / $\lambda$

$\lambda$ = ln(2) / T1//2

$\lambda$ = 0.693 / (5 x 24 x 3600)

$\lambda$ = 1.605 x 10^-6 s^-1

N = 125.8 x 10^4 Bq / 1.605 x 10^-6 s^-1

N = 7.84 x 10^11

b)

According to radioactive decay rate,

N = No*e^- $\lambda$ t

No = 7.84 x 10^11

t = 1 week = 7 days = 604800 sec

N =  7.84 x 10^11 * e^-(1.605 x 10^-6 s^-1 x 604800 )

N = 2.97 x 10^11