A radioactive sample has a count rate of 800 counts per minute. One hour later, the count rate has fallen to 100 counts per minute. What is the half-life of the sample?


Radioactivity equation states: N = No e^(-0.693t/t1/2) --- Where t = time, t1/2 = half-life, N = count after time t, No = Initial count Using the values given; 100 = 800 e^(-0.693*1/t1/2) 100/800 = e^(-0.693t/t1/2) 0.125 = e^(-0.693/t1/2) Taking natural logs on both sides; log 0.125 = -0.693/t1/2 * log (e) -0.9031 = -0.693/t1/2 * 0.4323 t1/2 = (-0.693*0.4323)/(-0.9031) = 0.3317 hours Therefore, half-life of the sample is approximately 0.3317 hours

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