How does this exponential decay formula get derived?

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Homework Statement:

Q12 e) A scan is taken over ts = 10.0 minutes. If the total energy detected by the scanner exceeds Emin = 5×10−4J, an image appears due to radioactivity in a thyroid gland. Approximately how much time will elapse before radioactivity is no longer detected in a scan?

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  • #2
etotheipi
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Here's my take on it; they have used ##A## for the initial activity, but I will redefine the notation so that ##A = A_0e^{-\lambda t}## is the activity as a function of time, and ##A_0## is the initial activity. If ##E_{\gamma}## is the energy of a single photon, then you have already worked out that the energy emitted per unit time is ##\dot{E} = E_{\gamma}A##.

If we inject at ##t=0## and wait until some time ##t## to start the scan, then we want to know what ##t## has to be in order for the radiation to just be able to be detected during a subsequent interval ##t_s##. That is, the total energy emitted between ##t## and ##t + t_s## is ##E_{\text{min}}##. Since ##A = A_0 e^{-\lambda t}##, this is expressed as$$E_{\text{min}} = \int_{t}^{t+t_s}E_{\gamma} A dt = E_{\gamma} A_0 \int^{t+t_s} e^{-\lambda t} dt = \frac{E_{\gamma} A_0}{-\lambda} \left[e^{-\lambda t} \right]_t^{t+t_s} = \frac{E_{\gamma} A_0}{-\lambda} (e^{-\lambda t}(e^{-\lambda t_s} - 1))$$ $$e^{-\lambda t} = \frac{\lambda E_{\text{min}}}{E_{\gamma} A_0 (1-e^{-\lambda t_s})}$$Now we have to use that ##e^x \approx 1 + x##, i.e. that ##1-e^{-\lambda t_s} \approx \lambda t_s##, which implies that$$e^{-\lambda t} \approx \frac{E_{\text{min}}}{E_{\gamma} A_0 t_s}$$ $$\lambda t \approx - \ln{\frac{E_{\text{min}}}{E_{\gamma} A_0 t_s}}$$ $$t \approx \frac{1}{\lambda} \ln{\frac{E_{\gamma}A_0 t_s}{E_{\text{min}}}} = \frac{t_{1/2}}{\ln{2}} \ln{\frac{E_{\gamma}A_0 t_s}{E_{\text{min}}}}$$Although this is exactly their expression, the answer is in fact approximate and not exact, as they claimed. Maybe they had something else in mind that I missed?
 
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  • #3
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Here's my take on it; they have used ##A## for the initial activity, but I will redefine the notation so that ##A = A_0e^{-\lambda t}## is the activity as a function of time, and ##A_0## is the initial activity. If ##E_{\gamma}## is the energy of a single photon, then you have already worked out that the energy emitted per unit time is ##\dot{E} = E_{\gamma}A##.

If we inject at ##t=0## and wait until some time ##t## to start the scan, then we want to know what ##t## has to be in order for the radiation to just be able to be detected during a subsequent interval ##t_s##. That is, the total energy emitted between ##t## and ##t + t_s## is ##E_{min}##. Since ##A = A_0 e^{-\lambda t}##, this is expressed as

$$E_{min} = \int_{t}^{t+t_s}E_{\gamma} A dt = E_{\gamma} A_0 \int^{t+t_s} e^{-\lambda t} dt = \frac{E_{\gamma} A_0}{-\lambda} \left[e^{-\lambda t} \right]_t^{t+t_s} = \frac{E_{\gamma} A_0}{-\lambda} (e^{-\lambda t}(e^{-\lambda t_s} - 1))$$ $$e^{-\lambda t} = \frac{\lambda E_{min}}{E_{\gamma} A_0 (1-e^{-\lambda t_s})}$$Now we have to use that ##e^x \approx 1 + x##, i.e. that ##1-e^{-\lambda t_s} \approx \lambda t_s##, which implies that$$e^{-\lambda t} \approx \frac{E_{min}}{E_{\gamma} A_0 t_s}$$ $$\lambda t \approx - \ln{\frac{E_{min}}{E_{\gamma} A_0 t_s}}$$ $$t \approx \frac{1}{\lambda} \ln{\frac{E_{\gamma}A_0 t_s}{E_{min}}} = \frac{t_{1/2}}{\ln{2}} \ln{\frac{E_{\gamma}A_0 t_s}{E_{min}}}$$Now this is an approximate answer; perhaps they had something else in mind that I missed?
ngl, Australian olympiad assumes that the student only knows algebra LOL. Pretty sure something is being missed.
 
  • #4
etotheipi
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ngl, Australian olympiad assumes that the student only knows algebra LOL. Pretty sure something is being missed.
Yes, in the solution manual they just expect you to take the activity to be approximately constant over the interval ##t_s##, in which case it is just algebra.

If you want the more accurate expression, I don't think there's any way of getting out of doing calculus. But they do not expect you to do this, they will have just added it as a comment in the mark scheme so that you can check how valid this approximation above is.
 
  • #5
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Yes, in the solution manual they just expect you to take the activity to be approximately constant over the interval ##t_s##, in which case it is just algebra.

If you want the more accurate expression, I don't think there's any way of getting out of doing calculus. But they do not expect you to do this, they will have just added it as a comment in the mark scheme so that you can check how valid this approximation above is.
Why does lambda equate to that term? That's the main part that I don't quite get.
 
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  • #6
etotheipi
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Why does lambda equate to that term? That's the main part that I don't quite get.
Ah, I apologise, I had not defined that. ##\lambda## is conventionally used to denote the decay constant, which is defined such that$$\lambda = \frac{\ln{2}}{t_{1/2}}$$It is the probability that a particular nucleus will decay per time ##dt##, which is why the activity (number of decays per second) goes as$$-\frac{dN}{dt} = A = \lambda N$$where ##N## is the total number of remaining nuclei. That allows you to derive the operational equations for radioactive decay, ##N = N_0 e^{-\lambda t}## or similarly ##A = A_0 e^{-\lambda t}##.

The reciprocal of ##\lambda## is called the lifetime, ##\tau = 1/\lambda##. This is not to be confused with the half life ##t_{1/2}##, which is a different thing!
 
  • #7
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Ah, I apologise, I had not defined that. ##\lambda## is conventionally used to denote the decay constant, which is defined such that$$\lambda = \frac{\ln{2}}{t_{1/2}}$$It is the probability that a particular nucleus will decay per time ##dt##, which is why the activity (number of decays per second) goes as$$-\frac{dN}{dt} = A = \lambda N$$where ##N## is the total number of remaining nuclei. That allows you to derive the operational equations for radioactive decay, ##N = N_0 e^{-\lambda t}## or similarly ##A = A_0 e^{-\lambda t}##.

The reciprocal of ##\lambda## is called the lifetime, ##\tau = 1/\lambda##. This is not to be confused with the half life ##t_{1/2}##, which is a different thing!
What is a qualitative way of understanding $$\lambda = \frac{\ln{2}}{t_{1/2}}$$?
 
  • #8
etotheipi
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I'm not sure there's a qualitative way. You start off with the probabilistic statement that ##-\frac{dN}{dt} = \lambda N##, which as I mentioned lets you derive ##N = N_0 e^{-\lambda t}##. That is achieved via. integration, but it is perfectly fine to just accept it for now.

Now you ask, how long will it take for the number of nuclei remaining in the sample to have halved? That is, initially ##N = N_0##, and at some later time which we'll call ##t_{1/2}## the total number of remaining nuclei is ##N = N_0/2##. Let's put this into the decay formula,$$\frac{N_0}{2} = N_0 e^{-\lambda t_{1/2}}$$Cancel the ##N_0##'s and take the natural logarithm of both sides,$$\ln{\left(\frac{1}{2} \right)} = -\lambda t_{1/2}$$$$-\ln{2} = -\lambda t_{1/2} \implies \ln(2) = \lambda t_{1/2}$$Which now tells you how the decay constant relates to the half life!
 
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