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What is the decay constant for a radioactive isotope which decreases to 34% of its original value in 2.48 hours?
15 years ago

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DrBob222
ln(No/N) = kt
ln(100/34) = k(2.48)
I started with 100 arbitrarily, then made the N value 34% of that. Solve for k.
15 years ago
me
thank you so much
15 years ago

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