Question

0.0100 uCi source of S-35 (t_1/2= 87.9 days) was counted in a liquid scintillation counter for 200 days. It gave a count of 1176 cpm. Calculate the percentage efficiency of this counting system .

Answer 1

The radioactive decay follows first order kinetics.

Let the initial activity (I_o) of the source S-35

=

0.0100μ,

Then, the activity (I) after time t = 200 days would be given by:

0.693 I = lo exp(- -xt)

The half-life t1/2 is = 87.9 days

and    t    = 200 days

which gives,

1= 10 x 0.2066

1 = 0.0021με,

Now, From the definition of Curie, one Curie is 3.7 × 10¹⁰ disintegrations per second (dps) or 2.22 X 10¹² disintegrations per minute (dpm)

So, for 0.0021 x 10 ^-6 Curie we have, (0.0021 x 10 ^-6 x 2.22 X 10¹² ) disintegrations per minute (dpm) = 4662 dpm

These are the disintegrations happening in all directions. The disintegrations counted by the instrument are given by cpm i.e. counts per minute.

The efficiency therefore = (cpm) / (dpm)

                                         = (1176) / (4662)

                                         = 0.2522

or, 25.22 % ... Answer

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