I've been given a problem but really no information. i used the half life formula and I would like someone to check my work please.
using the half life formula 1/2=A0e^kt
i did this
ln(1/2)=19.5t divide both sides by 19.5 to get t and came up with -0.03554600926
the only information i was given was the half life is 19.5 days and to find the decay rate k.
using the half life formula 1/2=A0e^kt
i did this
ln(1/2)=19.5t divide both sides by 19.5 to get t and came up with -0.03554600926
the only information i was given was the half life is 19.5 days and to find the decay rate k.
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The half life formula is just a little different, I think. You got the k value right, I think, except that I prefer to see the sign as positive (explain that later.) But the result looks good.
I think the half life formula is written more like this:
1. P(t) = P0•e^(-k•t)
Where in your case, P(t) = ½•P0 because you want the final population to be half as large at time t.
That's just a wee bit different that you show in your question. There's a minus sign there (which makes the _decaying_ nature manifest instead of hiding it into the constant, k) and there is a missing factor on the left side once you replace P(t) before solving. So I think you probably just didn't write it out correctly in your question, but still did things right afterwards.
So here is my work, keeping in mind that k is going to depend upon the units you decide for t. If t is in days, then the k will be in 1/day units. If the t is in years, then the k will be in 1/year units. Etc. So let's go with days, for now:
2. ½•P0 = P0•e^(-k•19.5)
3. ½ = e^(-k•19.5)
4. ln ½ = -k•19.5
5. k = -ln(½) / 19.5 = 0.0355460093
So I like your result. Except I don't like the sign. Your version makes it look as though this should be a rising exponential rather than a falling one. Placing the minus sign explicitly in the equation makes it clearer what is going on.
By the way, another way to state the equation is this:
6. P(t) = P0•e^(-k/τ)
In this case, τ (called tau) would be in days instead of 1/days. A lot of people prefer that form. In this case, of course, τ = 1/k = 28.1325533. Which is to say, if the population goes to ½ in 19.5 days, how many days does it take to go to 1/e? 28 days, or so. Putting it this way makes more sense to people (stating the constant in terms of days) than some oddball tiny factor that doesn't seem well related to the stated half-life.
I think the half life formula is written more like this:
1. P(t) = P0•e^(-k•t)
Where in your case, P(t) = ½•P0 because you want the final population to be half as large at time t.
That's just a wee bit different that you show in your question. There's a minus sign there (which makes the _decaying_ nature manifest instead of hiding it into the constant, k) and there is a missing factor on the left side once you replace P(t) before solving. So I think you probably just didn't write it out correctly in your question, but still did things right afterwards.
So here is my work, keeping in mind that k is going to depend upon the units you decide for t. If t is in days, then the k will be in 1/day units. If the t is in years, then the k will be in 1/year units. Etc. So let's go with days, for now:
2. ½•P0 = P0•e^(-k•19.5)
3. ½ = e^(-k•19.5)
4. ln ½ = -k•19.5
5. k = -ln(½) / 19.5 = 0.0355460093
So I like your result. Except I don't like the sign. Your version makes it look as though this should be a rising exponential rather than a falling one. Placing the minus sign explicitly in the equation makes it clearer what is going on.
By the way, another way to state the equation is this:
6. P(t) = P0•e^(-k/τ)
In this case, τ (called tau) would be in days instead of 1/days. A lot of people prefer that form. In this case, of course, τ = 1/k = 28.1325533. Which is to say, if the population goes to ½ in 19.5 days, how many days does it take to go to 1/e? 28 days, or so. Putting it this way makes more sense to people (stating the constant in terms of days) than some oddball tiny factor that doesn't seem well related to the stated half-life.
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Hint: Solve 1/2 = e^k for k,
k = ln(1/2)
Attn: To get decay rate, set t = 1.
k = ln(1/2)
Attn: To get decay rate, set t = 1.