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The isotope 2712Mg has a half-life of 9.46 min. Determine the time interval (in min) required...

The isotope 2712Mg has a half-life of 9.46 min. Determine the time interval (in min) required for the activity of a sample of this isotope to decrease to 37.0% of its original value.

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Answer #1

Half life of 2712Mg = T1/2 = 9.46 min

Decay constant =

= 7.327 x 10-2 min-1

Initial activity of the sample of the isotope = N0

Activity of sample left = N = 0.37N0

Time taken for the activity of the sample to be reduced to 37% of original = T

Taking natural log on both sides,

T = 13.57 min

Time taken for the activity of the sample to be reduced to 37% of original = 13.57 min

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