Question

For an exposure of 43 R (Roentgen), calculate;

a) The charge (in coulombs) liberated per kg of air

b) The number of ion pairs liberated per kg of air

c) The energy absorbed per kg of air

d) The absorbed dose in air

Answer 1

The question asked is about Dosimetry.

a) Roentgen is a unit of exposure and named after Wilhelm Roentgen . He discovered in the year 1895 , the x rays. The definition of 1 R(Roentgen) in terms of Coulombs per kilogram is 2.58 * 10 ^-4 coulombs per kilogram.(C kg^-1)

So,

1 R(Roentgen) = 2.58 * 10 ^-4 (C kg^-1)

43 R (Roentgen) = 43 * 2.58 * 10 ^-4 (C kg^-1)

= 110.94 * 10^-4 (C kg ^-1)

= 0.011094 (C kg^-1)

hence the charge (in coulombs) liberated per kg of air is 0.011094 (C kg^-1)

b) As per definition the no. of ion pairs per Coulomb in air are given by 1.6*10¹⁹ ion pairs per coulomb (IP C^-1)

We already know that in 1R (Roentgen) there is 2.58 * 10 ^-4 coulombs per kilogram.

So no. of ion pairs in 1R = ( 2.58 * 10 ^-4 (C kg^-1)) * (1.6*10¹⁹  (IP C^-1))

= 2.58 * 10 ^-4 * 1.6*10¹⁹

= 4.128 * 10¹⁵ ( IP kg^-1)

Therefore no. of ion pairs in 43 R = 4.128 * 10¹⁵ ( IP kg^-1) * 43

=177.504 * 10¹⁵ (IP kg^-1)

Thus the number of ion pairs liberated per KG of air is 177.504 * 10¹⁵

c) Energy absorbed in air per ion pair is given as 33.7 electron-volt per ionpair (ev IP^-1)

We know , no. of ion pairs in 1R is 4.128 * 10¹⁵ ................(from b part )

therefore energy absorbed in 1R = (4.128 * 10¹⁵ (IP kg^-1 )) * (33.7 (ev IP^-1))

= 4.128 * 10¹⁵ * 33.7

=139.1 * 10¹⁵ (ev kg^-1)

Therefore energy absorbed in 43 R = 139.1 * 10¹⁵ * 43 (ev kg^-1)

= 5981.3 * 10¹⁵

to convert this measure in Joules we will divide it by (1.6 * 10 ¹⁹), since there are 1.6 * 10 ¹⁹ electron volts(ev) in 1 Joule. thus in joules the energy is

3738 *10^-4 Joules

Thus energy absorbed per Kg of air is 5981.3 * 10¹⁵ ev or 3738 *10^-4 Joules

d) There are 1.6 * 10 ¹⁹ electron volts(ev) in 1 Joule. and the energy absorbed in 1R is 139.1 * 10¹⁵.

thus the absorbed dose in air in 1R = 139.1 * 10¹⁵. / 1.6 * 10 ¹⁹

= 86.9 * 10^-4 (joules per Kg)

= 0.00869 (joules per kg) or (gray)

One gray equals 100 rads.

Thus an exposure of 1 R (roentgen ) = 0.00869 * 100 of absorbed dose

= 0.869 rads of absorbed dose

Thus an exposure of 43R (roentgen) = 0.869 * 43 rads of absorbed Dose

= 37. 367 rads of absorbed dose.

Thus the absorbed dose in air is 37.367 rads.

If you find the answers as stated above please upvote and comment. Thank you. Cheers