The half life of sulfur-38 is 2.87 hours. after 8.61 hours, what would the percent of the original radiation is left?

12.5% remaining after 8.61 hrs or 3 half-lives.

Explanation:

Method I: 1st Order Decay Equation

Given t(1/2) = 2.87 hrs => k = 0.693/t(1/2) = (0.693/2.87) hr⁻¹ = 0.242 hr⁻¹

C(final) = C(initial)e^-k·t => ln(C/C₀) = -k·t => lnC - lnC₀ = - k·t => lnC = lnC₀ - k·t

lnC = ln(100%) - [(0.242 hr⁻¹)(8.63 hr)] = (4.605 - 2.09)% = 2.517%

lnC = 2.517% => C = (e²°⁵¹⁷)% = 12.5% remaining

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Method 2 - Decay Trend Graph => see attachment

Note: All radio decay trends follow 1st order kinetics. A tracing of the decay trend for a 1st order process with half-life of 2.87 hrs shows S-38 decays over 3 half-lives (= 8.61 hrs). Therefore ...

after 1st half-life => 50% remains

after 2nd half-life => 25% remains

after 3rd half-life => 12.5% remains

it means there are 3 atoms of oxygen