Question

¹⁸F (half-life of 110 min) is produced at a rate of 1 Ci/hr per one gram of solution.

1, Calculate the activity (in Ci) resulting after irradiating 10 grams of solution for one hour.

2, Calculate the activity (in Ci) resulting for an irradiation of 2 gm solution for 5 hours

3, Calculate the activity (in Ci) resulting for an irradiation of 1 gm solution for 10 hours

Answer 1

A_o is the rate of production of the activiity

is the decay const ( = 0.693/T_1/2 )

at any instant of time A_o nuclie are produced, and if there are N(t) nuclie present,

out of this N(t) would decay

the rate of change of nuclie is given by

dN(t)/dt = A_o - N(t)

intially the activity is 0 integrating the above from 0 to t

the activity of the isotope at any time is given by

N(t) = A_o {1- exp(-t) }

= 0.693/ T_1/2 = 0.693/110 = 0.0063 m^-1 = 1.05e-4 hr^-1  

production rate A_o = 1Ci/hr/gm

1. 10 gms solution A_o = 10Ci/hr , irradiation time t = 1hr

acitivity after 1 hr = 10(1-exp(-1.05e-4) ) = 1.05e-3 Ci

2. 2 gms solution A_o = 2Ci/hr ; irradiation time t = 5hr

acitivity after 2 hrs = 2(1-exp(-5.25e-4) ) = 1.05e-3 Ci

3. 1 gm solution, A_o = 1Ci/hr ; irradiation time t = 10hr

acitivity after 10 hrs = 1(1-exp(-1.05e-3) ) =1.05e-3 Ci