Aradioactive 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

Ithink your answer is d