How does this exponential decay formula get derived?

In summary, the conversation discusses a formula for calculating the time needed for radiation to be detected during a scan, given the initial activity and energy of a single photon. The formula is derived using calculus, but the solution manual simplifies it by assuming constant activity. The use of lambda in the formula is due to the activity function, which is expressed as A = A0e^-lambda*t.
  • #1
aspodkfpo
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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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https://www.asi.edu.au/wp-content/uploads/2015/03/PhysicsASOE2013soln.pdf
Q12 e)

Working backwards, P = Ae^kt form, i.e. EAts = Emin e^(ln2/τ x t).

Not sure how they get this formula in the first place with these values.
 
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  • #2
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
etotheipi said:
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
aspodkfpo said:
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
etotheipi said:
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
aspodkfpo said:
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
etotheipi said:
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
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 let's 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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1. What is the exponential decay formula?

The exponential decay formula is a mathematical expression that describes the decrease in value of a quantity over time. It is typically written as y = ae-bx, where a and b are constants, x is the independent variable (usually time), and y is the dependent variable (usually the quantity being measured).

2. How is the exponential decay formula derived?

The exponential decay formula is derived using calculus. It involves taking the derivative of the exponential function, which results in a negative constant multiplied by the original function. This negative constant is the rate of decay, and when integrated, it becomes the b constant in the formula.

3. What are the assumptions made in deriving the exponential decay formula?

There are a few assumptions that are made in deriving the exponential decay formula. These include a constant rate of decay, no external factors affecting the decay, and a continuous and infinitely divisible time interval.

4. Can the exponential decay formula be applied to any situation?

The exponential decay formula is most commonly used in situations where the quantity being measured decreases over time, such as radioactive decay or population decline. However, it can also be applied to other scenarios as long as the assumptions hold true.

5. How accurate is the exponential decay formula?

The accuracy of the exponential decay formula depends on how well the assumptions hold true in the given situation. In real-world scenarios, there may be external factors that affect the decay rate, making the formula less accurate. However, in controlled experiments where the assumptions are met, the formula can provide a very accurate prediction of the decay of a quantity over time.

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