Radioactivity of a sample decreases 90% after 10 years. What is the half-life of the sample?Note: Radioactivity follows first order kinetics.
Question
Radioactivity of a sample decreases 90% after 10 years. What is the half-life of the sample?Note: Radioactivity follows first order kinetics.
Solution
Sure, let's solve this step by step:
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First, we need to understand that the decay of radioactive substances follows an exponential decay law, which can be expressed as N = N0 * e^(-kt), where N is the final amount, N0 is the initial amount, k is the decay constant, and t is time.
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In this case, we know that the radioactivity decreases by 90% after 10 years. This means that N/N0 = 0.1 (because only 10% remains).
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We can substitute these values into the decay equation: 0.1 = e^(-k*10).
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To solve for k, we can take the natural logarithm of both sides: ln(0.1) = -k*10.
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Solving for k gives us k = -ln(0.1)/10.
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The half-life (T) of a radioactive substance is related to the decay constant by the equation T = ln(2)/k.
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Substituting the value of k we found, we get T = ln(2)/[-ln(0.1)/10].
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Simplifying this gives us T = -10*ln(2)/ln(0.1).
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Calculating this gives us a half-life of approximately 3.3 years.
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