Using the derivative of the formula of the number of nuclei

[Albo1125]

Hi all,

I have a question concerning the derivative of the formula of the number of nuclei. I hope I've posted this in the right section, I'm new here :P. Anyway, in the question, the given values are:
At a certain time t, there is an amount of radioactive Br-82. The activity A is 7.4*10¹⁴ Bq, the number of nuclei N is 9.6*10¹⁸.
I'm meant to calculate t_1/2, the time it takes before the number of nuclei has halved.

Homework Equations

N = N₀ * (½)^t/t_1/2
The derivative of the formula above (given by the coursebook, so it's correct): A = ( ln(2) * N) / t_1/2.

The Attempt at a Solution

What seemed like a plausible solution to me was to use the derivative to calculate t_1/2.
So: t_1/2 = ( ln(2) * N) / A.
That gave me: t_1/2 = ( ln(2) * 9.6*10¹⁸ ) / ( 7.4*10¹⁴) = 8992 seconds. This derivative always uses t_1/2 in seconds.
This is equivalent to 8992/3600 = 2.50 hours. So t_1/2 = 2.50 hours.
The correct answer, however, is 35.3 hours.

Can anybody explain what mistake I have made, or why I can't use the derivative in this situation?
Thank you very much in advance :)

 

[Fredrik]

When you write $t_\frac{1}{2}=\frac{N\ln 2}{A}$, you haven't actually solved for $t_\frac{1}{2}$, since N depends on $t_\frac{1}{2}$.

 

[Albo1125]

When you write $t_\frac{1}{2}=\frac{N\ln 2}{A}$, you haven't actually solved for $t_\frac{1}{2}$, since N depends on $t_\frac{1}{2}$.

By this, I assume you mean I cannot use the starting values of A and N in that formula? Because in a different question where I non-starting values for A and N, my method did actually work.

 

[Albo1125]

By this, I assume you mean I cannot use the starting values of A and N in that formula? Because in a different question where I had non-starting values for A and N, my method did actually work.

 

[Fredrik]

I'm going to write T instead of $t_\frac{1}{2}$ to reduce the amount of typing.

I've had a closer look at the problem now. At first I thought you needed to solve the equation $A(t)=\frac{N_0\cdot 2^{-t/T}\cdot\ln 2}{T}$ for $T$, but you don't. The problem gives you numerical values for A(t) and N(t) at one specific (but unspecified) time t. At that time, we have
$$A(t)=\frac{N(t)\ln 2}{T}.$$ So I see nothing wrong with the result
$$T=\frac{N(t)\ln 2}{A(t)}.$$ I will think about this some more, but right now I see nothing wrong with what you did. The result will be in seconds, because N(t) is given as a number of nuclei, and A(t) is given as a number of nuclei per second.

 

[Albo1125]

I'm going to write T instead of $t_\frac{1}{2}$ to reduce the amount of typing.

I've had a closer look at the problem now. At first I thought you needed to solve the equation $A(t)=\frac{N_0\cdot 2^{-t/T}\cdot\ln 2}{T}$ for $T$, but you don't. The problem gives you numerical values for A(t) and N(t) at one specific (but unspecified) time t. At that time, we have
$$A(t)=\frac{N(t)\ln 2}{T}.$$ So I see nothing wrong with the result
$$T=\frac{N(t)\ln 2}{A(t)}.$$ I will think about this some more, but right now I see nothing wrong with what you did. The result will be in seconds, because N(t) is given as a number of nuclei, and A(t) is given as a number of nuclei per second.

Did you find out yet? :)

 

[Fredrik]

I still don't see anything wrong. The problem is saying that there's a real number t such that $N(t)=9.6\cdot 10^{18}$ and $A(t)=7.4\cdot 10^{14}$. Since
$$A(t)=-N'(t)=-\frac{d}{dt}\left(N_0\cdot 2^{-t/T}\right) =-N_0\cdot 2^{-t/T}(\ln 2)\left(-\frac{1}{T}\right) =\frac{N_0\cdot 2^{-t/T}\cdot \ln 2}{T} =\frac{N(t)\ln 2}{T},$$ this means that
$$T=\frac{N(t)\ln 2}{A(t)}=\frac{9.6\cdot 10^{18}\cdot \ln 2}{7.4\cdot 10^{14}} =\frac{9.6\ln 2}{7.4}\cdot 10^4\approx 8992.$$