Question

A 72 kg lab-worker recieves a dose of 22 mSv from a 0.95 MeV beta-emitter (Q=1) over the course of one hour.

A.) What is the dose in Gy?

B.) How much energy is absorbed by the lab-worker in J?

C.) How many beta particles does this correspond to?

Answer 1

ANSWER :-

A) 1 Gy = 6.24 x 1012 MeV /kg

Given 0.95 MeV and Mass of lab worker as 72 kg

Dose =[ ( 0.95 x 72 ) / 6.24 x 1012 ] Gy  

= 10.96 x 10-12 Gy

B) 1 Gy = 1 J/kg

therefore  10.96 x 10-12 Gy = 10.96 x 10-12  x 72 = 789.23 x 10-12  J

C) beta partcle corresponds to electrons

Mass of an electron = 9.10938 x 10-31 Kg

Let there are n number of electrons

Using E = n x mc2

Therefore , 789.23 x 10-12 = n x 9.10938 x 10-31 x[ 3 x108 ]2

n =  789.23 x 10-12 / 9.10938 x 10-31 x[ 3 x108 ]2

n = 9.627 x 103 / second