A 75 kg person receives a whole-body radiation dose of 24 m-rad. delivered by ∝-particles for which the RBE factor is 12. Calculate (a ) the absorbed energy in Joules, and (b ) the equivalent dose in rem.

[Ans : (a ) 18 m J (b ) 0.29 rem)



Given Data:

Mass of the body = mr₁ = 75 kg

Absorbed radiation dose = D = 24 m-rad = 24 ｘ 10⁻³ rad = 24 ｘ 10⁻³ï½˜ 0.01 Gy = 2.4 ｘ 10⁻⁴ Gy
[1 rad = 0.01 Gy]

RBE (Relative Biologic Effectiveness) Factor of radiation = 12




To Find:

The energy released (in Joule) = E = ?

The equivalent dose in rem = Dâ‚‘ = ?


Solution:


Since

D = `\frac {E}{m}`

or

E = Dm

Putting values

E = 2.4 ｘ 10⁻⁴ Gy ï½˜ 75 kg

E = 180 ｘ 10⁻⁴ J

E = 18.0 ｘ 10⁻³ J
 
E = 18.0  m J-----------------Ans. 1


Now we know that 

Dâ‚‘ = D ｘ RBE

Dâ‚‘ = 2.4 ｘ 10⁻⁴ Gy ï½˜ 12

Dâ‚‘ = 28.8 ｘ 10⁻⁴ Gy 

or

Dâ‚‘ = 2.88 ｘ 10⁻³ Gy 

As 1 Gy = 0.01 rem,

Dâ‚‘ = `\frac {2.88 ｘ 10⁻³ }{0.01}` rem

Dâ‚‘ = 0.288    rem ------------------Ans.2




************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.