Am-241 is a radioactive isotope that decays by alpha emission. Each alpha partic

Question

Am-241 is a radioactive isotope that decays by alpha emission. Each alpha particle emitted carries 8.79E-13 J of kinetic energy.

A 110.0 kg worker is accidentally exposed to a sample of Am-241 having an activity of 0.0600 curies for a total of 3.00 minutes. During that time the workers body absorbs 10.00 percent of the alpha particles emitted by the sample.

How many alpha particles were emitted by the sample during the time of the workers exposure?
Answer alpha particles emitted

How many joules of energy were absorbed by the worker?
Answer Joules

The alpha particles emitted by the sample have an RBE=20. Calculate the workers exposure in millirems.

Explanation / Answer

Activity of Am-241 = 0.0600 Ci

Time of absorbtion of radiation : 3 min = 3 x 60 = 180 sec

1 Ci = 3.7 x 1010 dps (dps: disintegrations per second)

So 0.0600 Ci corresponds to 0.0600 Ci x 3.7 x 1010 dps = 2.22x 109 dps

No. of alpha particles emitted in 3.00 minutes = 2.22x 109 x 180 =399.6x109 = 3.99x1011

3.99x1011 alpha particles were emitted by the sample during the time of the workers exposure

Energy of one alpha particle = 8.79x10-13 J

Percentage absorption of alpha particles by the human body = 10%

So the energy absorbed due to alpha particles is : (10/100) x 3.99x1011 x 8.79x10-13 J = 0.3507 J = 35.07 mJ

35.07 mJ of energy was absorbed by the worker

Exposure calculations:

RBE = 20

Rem = (Energy absorbed /Mass) x RBE

Rad = (35.07 mJ/ 110 kg) x 20 = 6.37 mSv

Dose in Rem = 6.37 x 100 = 637 rem

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