# Radioactivity of an unknown isotope

1. Aug 27, 2006

### khy86

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.

2. Aug 27, 2006

### HallsofIvy

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
$$N(t)= N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}}$$.

3. Aug 27, 2006

### Astronuc

Staff Emeritus
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.

Last edited: Aug 27, 2006
4. Aug 27, 2006

### khy86

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 begining 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, thats the reason why i assumed it was wrong.