Radioactivity of an unknown isotope

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The discussion centers on deriving a formula to calculate the remaining atoms of an unknown isotope based on its half-life. The original equation provided, delta(N) = N(t).delta(t)/T, leads to confusion regarding the meaning of delta(N) and the term "t.sq." Clarifications reveal that delta(N) represents the change in the number of atoms over a time interval, and the correct decay formula is N(t) = N0(1/2)^(t/T). The participants also note that the decay rate should be negative, indicating a decrease in the number of atoms over time. Concerns are raised about the formula yielding results greater than the initial atom count when tested in Excel, indicating a potential misunderstanding in the application of the equations.
khy86
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i have been given an equation of,
delta(N) = N(t).delta(t)/T

where N(t) is the number of atoms left at time t, and T is the half life. Using a constant time and variable half lifes i have to come up with a formula to calculate the remaining atoms left over.

the formula i was able to obtain from my working is

delta(N) = N0/e^(t.sq) / 2T

i am not sure if this equation is write and also my problem is that i have no idea of how i obtained this formula from the original one.
 
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You tell us what N(t), t, and T are but you don't tell us what delta(N) means. Is it the change in N during the time interval delta(t)? In the resulting formula I do not understand what you mean by "t.sq" nor what you mean by A/B/C. Is that the same as A/(BC) (A divided by B, then that divided by C)?

"i have no idea of how i obtained this formula from the original one."
Well, you were the one who obtained it! What did you do?

Assuming that delta(N) is indeed the change in N during time interval delta(t), then, dividing by delta(t) and taking the limit as delta(t) goes to 0, \frac{dN}{dt}= N/T, a differential equation. We then have
\frac{dN}{N}= \frac{dt}{T}
where T is a constant (the half life). Integrating both sides,
ln(N)= t/T+ C or
N(t)= Ce^{\frac{t}{T}}.
Taking t= 0, N(0)= N0, that gives
N(0)= N_0= Ce^0= C
so
N(t)= N_0 e^{\frac{t}{T}}

My assumption about what "delta(N)" means must not be correct then, because that is not correct!

If T is the "half life", then for every period of length T, the number of atoms left is cut in half. In particular, the "number of periods of length T in time t" is just \frac{t}{T} so we multiply the initial amount, N0 by 1/2 \frac{t}{T} times. That is
N(t)= N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}.
 
Normally the rate of decay is expressed as

dN(t)/dt = -\lambda N(t), i.e. the decay rate is directly proportional to the number of atoms, N, at time t, and is decreasing.

\lambda is the decay constant and = (ln 2)/T1/2, where T1/2 is the half-life.
 
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you are right, delta(N) is the change, sorry my bad for not explaining, i wasnt sure how to put the triangle in. Anyway the formula you obtained being N(t) = N0.e^(t/T). Using my own solving methods i obtained that but assumed it was wrong.

When i put the equation into excel, and put in some random values of t, N and T. the final value of N(t) ended up being higher than the beginning value of the number of atoms.

For example, the N being 10, t being 1 and T being 10, the final value ended up being 11.05, that's the reason why i assumed it was wrong.
 
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