Locked SIM 1007 PM Assignment 2doc black boardbsuedu uring a

Locked SIM 10:07 PM Assignment 2.doc black board.bsu.edu uring a leaN m Lane raamaulom sane set up. Find the equivalent dose H, if the absorbed is received through (i) y-rays of energy 15 MeV (ii neutrons of energy 12.5 MeV (iii) a particles ofenergy 4.2 MeV. (b) An equivalent dose of 45 Sv is recommended in a certain application of radiations. How much absorbed dose should be given to achieve the goal if t has to be supplied through electrons of energy 5 MeV (ii) neutrons of energy 5 MeV (iii protons of 5) (iv) a-rays of energy 3 MeV. energy 10 MeV (W Three organs of the body of a classified radiation worker is exposed to radiation in one year period. His Lungs receive equivalent dose of 250 mSv. Gonads receives 120 mSv of equivalent dose, while his thyroid gets 300 mSv of equivalent dose. This dose exceeds the annual limit of 50mSv for him. He can control this situation only by hiding his lungs a little bit to receive little dose over there. By how much he should decrease the equivalent dose given to lungs so that the total effective dose he receives in one year is 50 mSv? 4. (a) X-rays are allowed to prod uce 3 x 10i ion pairs in air of mass 500 grams. How many Roentgen of exposure is produced? (b) Calculate the quantity of charge produced as a result of 4 RExposure in dry air of mass 0.008 mg. If this charg is e the result of electrons production then find the number of electrons produced.

Solution

weight factor for Lungs = 0.12

                           Thyroid = 0.04

                          Gonads = 0.08

Total effective dose = 250 *012 +300*0.04 + 120 *0.08 = 51.6 mSv

dose excceded by 1.6 mSv

This has to be decreased from dose to Lungs

Lungs dose has to come down by 1.6/0.12 = 13.33 mSv

4. 1 Rontgen is equal to exposed does that liberates 2.58e-4 C /kg pf air

ions pars produced = 3.0 e+14 in 500 gms of air

    charge produced per kg = 3.0e+14 *2 *2 *1.6e-19 C = 19.2 e-5 C /kg

Exposure dose                      =    19.2 e-5 / 2.58e-4 = 0.744 R

b) dose 4R

charge produced = 4 *2.58e-4 C/kg

mass of air = 0.008 mg

charge produced in the medium = 4*2.58e-4 *0.008e-6 C

                                           = 4.128e-13 C

        number of electrons liberated = 4.128e-13 /1.6e-19 = 2.58e+6