Find half life of radioactive subst. reduced by 36% in 38 hr
- Thread starter kalpana
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Re: pre calculus
You don't need the initial value. Put in a general initial value, like A, and then work the steps. The half-life does not depend on the initial value, since it is expressed as the percentage that is reduced in a fixed amount of time.
You don't need the initial value. Put in a general initial value, like A, and then work the steps. The half-life does not depend on the initial value, since it is expressed as the percentage that is reduced in a fixed amount of time.
mmm4444bot
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Re: pre calculus
Do you realize that there are two steps to this exercise? You need to find k before you can find t. Use the same formula in your other post:
P = Po e^(kt)
Here's a hint (for the first step) to go with the previous reply by Roy.
If you choose to use Po = A, then you will have 0.64A after 38 hours
My edit: Corrected wrong percentage (Thank you, Galactus)
Do you realize that there are two steps to this exercise? You need to find k before you can find t. Use the same formula in your other post:
P = Po e^(kt)
Here's a hint (for the first step) to go with the previous reply by Roy.
If you choose to use Po = A, then you will have 0.64A after 38 hours
My edit: Corrected wrong percentage (Thank you, Galactus)
If the substance reduces 36% in 38 hours, then there is 64% remaining after 38 hours.
A handy formula to know for half-life is \(\displaystyle T=\frac{-1}{k}ln(2)\)
All you need is the decay rate k.
You can find k by using \(\displaystyle .64P=Pe^{k(38)}\)
Solve for k.
A handy formula to know for half-life is \(\displaystyle T=\frac{-1}{k}ln(2)\)
All you need is the decay rate k.
You can find k by using \(\displaystyle .64P=Pe^{k(38)}\)
Solve for k.