Question

40. A sample consists of \(1.00 X 10^6\) radioactive nuclei with a half-life of 10.0 h. No other nuclei are present at time t = 0. The stable daughter nuclei accumulate in the sample as time goes on. (a) Derive an equation giving the number of daughter nuclei \(N_d\) as a function of time. (b) Sketch or describe a graph of the number of daughter nuclei as a function of time. (c) What are the maximum and minimum numbers of daughter nuclei, and when do they occur? (d) What are the maximum and minimum rates of change in the number of daughter nuclei, and when do they occur?

Answer 1

@theEric
@Saeeddiscover

Answer 2

I got an apparent equation, but when I attempted to find the minimum and maximum in part c (I skipped b) and took a derivative, I then tried setting it equal to 0 but it's impossible to solve for "t" because that requires taking the natural log of 0 which makes no sense.

The minimum and maximum of the rate implies the same thing since the derivative of \(e^{ax}\) is simply \(a*e^{ax}\)

Answer 3

I don't know nuclear physics, sorry! Good luck!

Answer 4

The problem really isn't the nuclear physics, it's the math behind it. The math makes no sense

Answer 5

Also, radioactivity and half-life are a part of classical physics, not nuclear physics.
\[N_0 = 1.00 X10^6 nuclei\]
\[T_{1/2}=10 hours\]
\[\huge N(t)=N_0e^{-\lambda t}\]
\[\huge N_d=N_0-N\]

So by substituting the numbers, that's the equation I get for \(N_d\), but it's not something you can differentiate to get the max/min of.

Answer 6

\[\huge \lambda = {ln2\over {T_{1/2}}}\]

Answer 7

If you want the
"maximum and minimum rates of change in the number of daughter nuclei,"
Then you'll need to find
\(\dfrac{\text dN_d}{\text dt}\) to get the rate of change in \(N_d\)

and you'll need to find
\(\dfrac{\text d^2N_d}{\text dt^2}\) so you can set it to zero to find the minimums and maximums.

So you have
\(\large N_d=N_0-N\)
so substituing for \(N(t)\)
\(\large N_d=N_0-N_0e^{-\lambda t}\)
That is a function of \(t\) you can differentiate many times.

Answer 8

So that's for part (d), right?

Answer 9

While I agree with your method, there's just one problem, you cannot solve for t because you cannot take the natural log of 0

Answer 10

Are we trying to solve for \(t\)?

Answer 11

Well, t is the only variable....so I'm not sure....

Answer 12

\(\Large\dfrac{\text dN_d}{\text dt}=\dfrac{\text d}{\text dt}\left(N_0-N_0e^{-\lambda t}\right)\)

Answer 13

ok, right, that's the derivative I want to take.

Answer 14

And that would be
\(\large\dfrac{\text d}{\text dt}\left(N_0-N_0e^{-\lambda t}\right)=N_0\lambda \ e^{-\lambda t}\)

Answer 15

I think. :)

Answer 16

N_0 is a constant so it becomes 0. The decay constant lambda comes down.
Since I'm looking for max and min, I want to set that equal to 0 right?

Answer 17

So if I set it equal to 0, the only variable I an solve for is t, which is impossible

Answer 18

That derivative is (by defination) the rate of change of the number of daughter nuclei. You want to take the derivative again, and set that equal to zero.

Answer 19

I still get the same problem though...
Since that is an exponential, the only thing I'm doing is bringing down a constant. The exponential does not change.

Answer 20

You have the rate of change, right?

The minimums and maximums of a function exist where the rate of change of that function is zero.

So we want the rate of change of the rate of change to be zero.
\(\dfrac{\text d^2N_d}{\text dt^2}=-N_0\lambda^2\ e^{-\lambda t}\)

\(-N_0\lambda^2\ e^{-\lambda t}=0\)
\(\Downarrow\) assuming \(N_0,\lambda\neq0\)
\(e^{-\lambda t}=0\)

Answer 21

I see....

Answer 22

You see my problem now? Natural log of 0.

Answer 23

Haha, \(-\infty\).

Answer 24

See, exponential curves don't have minima or maxima.

Answer 25

So then what is the problem asking? Under that assumption, isn't it asking for something that is impossible?

Answer 26

In fact, the greatest rate of change would be as \(t\rightarrow-\infty\)

Answer 27

Is the equation for \(N_d(t)\) correct for sure?

Answer 28

Yes. It is asking for daughter nuclei based on parent nuclei meaning that as time passes, the parent decays into the daughter but the sum of parent and daughter still equal the parent. It is the conservation of charge.

Answer 29

There might be some physics property to help.

But the rate of change in numbers of daughter nuclei is maximum at \(\infty\), where \(t\rightarrow-\infty\).

It is a minimum at about \(0\), where \(t\rightarrow\infty\).

Does that count?

Answer 30

Would that be how exponential decay works?

Answer 31

In all honesty, by book has a numerical answer but it says:

The number of daughter nuclei first increases most rapidly, at 6.93 X10^4/h, and then more slowly. It's rate of change approaches zero in the far future.

So I guess part of it goes to infinity.

Answer 32

This particular exponential decay is a representation of half-life. I'm sure you're familiar with the concept of carbon dating?

Answer 33

Oh, whoops, I meant "my book".

Answer 34

A little bit. I know the concept behind half-life. It is the length of time necessary to reduce an amount to its half.

Answer 35

Haha, I read "my book" anyway. It's late :P

Answer 36

I know where the 693 comes from but not the factor of 10.
It's coming from the ln2.

Answer 37

Interesting.
Is \(T\) the period?

So the answer is in the unit \(\rm/h\)?

Answer 38

which T are we talking about?

Answer 39

The \(T\) in \(\Large\lambda = {\ln2\over {T_{1/2}}}\)

Answer 40

Then \(\lambda\) has the dimension of inverse time like the answer in your book.

Answer 41

okay, that T is literally the half-life
for example the half-life of carbon-14 is 5730 years.
so after the allotted time, half of the carbon 14 will decay into Nitrogen 14 an electron, and an anti-neutrino through negative beta decay

Answer 42

Haha, I don't know why I said period, thanks!

Answer 43

This may clear up some of the confusion in symbols

Answer 44

I didn't read that all though before, sorry.

Answer 45

what time is it where you are @theEric ?

Answer 46

3:03 a.m. What is it for you?

Answer 47

12:03am; so you're on the east coast; I'm in CA
But regardless, it is late, or early depending on how you look at it.
Want to call it a night (or morning)?

Answer 48

I'm totally lost with this thing.

Answer 49

I might soon! This is material I haven't encountered yet, but it's always nice to get ahead.

Answer 50

You're a student?

Answer 51

That's somewhat surprising to hear.

Answer 52

Yep! Undergraduate in college in a physics program for a Bachelor's degree.

Answer 53

Heh, so am I actually. Well anyways, I'm gonna call it a night, you really should do the same. :)

Answer 54

Thanks for the attempt though.

Answer 55

So it's not too odd!
I won't be on as much tomorrow. I have more homework and some plans.

Answer 56

My pleasure! I wish you luck with this!

Answer 57

So, is \(N(t)\) the decay rate, like \(R\) in the attachment you posted?

Answer 58

To form your equation,
\(N_0=N+N_d\)
\(\implies N=N_0-N_d\)
I thought \(N_0\) is the total amount, \(N\) is the undecayed amount, and \(N_d\) is the decayed amount, so there is this conservation.

But the attachment says that the formula that you have for \(N(t)\) is for the decay rate. This wouldn't change the exponential, though.

\(\left|\dfrac{\text dN}{\text dt}\right|=R_0e^{-\lambda t}\)
\(\implies N+c=\int R_0e^{-\lambda t}\text dt\)
\(\implies N=-\dfrac{R_0}{\lambda}e^{-\lambda t}+c\)

Answer 59

I should have wrote \(N_d=N_0-N\) at the top of my last reply.

Answer 60

It still gives the maximum as \(\lambda R_0e^{-\lambda t}=0\), in which the exponential is still there.....

It's odd...

Good luck! :)