Asked Nov 14, 2019

The half life of 52Fe is 8.28 hours. Suppose we measure the activity of a sample containing 52Fe and find it to be 666 disintegrtions per minute. If we measure its activity the next day (say after 24.7 hours), what will its activity be at this point?


Expert Answer

Step 1

Calculation of disintegration constant (λ):

T1/2 of 52Fe is = 8.28 hours.


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0.693 disintegration constant (A)= T. 0.693 = 2.3248x10 s 8.26x3600

Step 2

Converting the given radioactivity into curie unit:

The given activity i...


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ldpm 4.5x101 Curie(or)Ci 666dpm 666x 4.5x1013 A,=2997x1010 Ci


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