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Homework Help: Radiation decay

  1. Nov 15, 2004 #1
    Let’s say you are irradiating a tumor that needs to receive a certain dose of energy. For the tumor to receive this radiation it must pass through the skin. Because it decays with r^2 the skin sees more radiation then the tumor and will always receive more (unless you use multiport not considered for this problem). The goal of minimizing this difference so as not to cause more cancer than you are treating lead to this statement: my prof claims that you can increase the distance of the patient from the source to minimize the radiation seen by the skin while still keeping the dose at the tumor site. He claims that as both r1 and r2 increase the difference become less important.

    I feel this is not correct because if you work back from the tumor the skin will always see that same amount of radiation. Which way of thought is correct?

    If anyone can follow that they are a guru of illogical thought at the very least.

  2. jcsd
  3. Nov 15, 2004 #2
    Are you saying that the incident power is inversely proportional to r^2? i.e.

    [tex] P = \frac{k}{r^2}[/tex]

    If so, consider a linear approximation of the difference at a distance r.

    [tex] \frac{dP}{dr} = \frac{-k}{2r^3}[/tex]
    [tex] dP = \frac{-k}{2r^3}dr[/tex]
    [tex] \Delta P = \frac{-k}{2r^3}\Delta r[/tex]

    Notice that the difference in power between the skin and the tumor decreases the larger the distance to the skin is, since the denominator increases as r increases.

  4. Nov 15, 2004 #3
    Regardless of how far away the source is the tumor must receive the same amount of energy. So if I work in reverse order does the radiation the skin is exposed to remain constant with any distance? This would be accomplished by just increaseing the output from the machine. The way it was explained is that the further out you are the "flatter the decay graph". But if you work backwards from the tumor the skin will always see the same amount of radiation? I feel this is flawed logic becasue the magnitudes will be different but the values at the two points I care about (the skin and tumor) must remain constant.
  5. Nov 16, 2004 #4
    The tumor receives the same amount of energy, regardless, yes. However, the skin receives more energy than the tumor. The amount more depends on the distance from the source.

    Suppose the distance to the skin is r and the distance from the skin to the tumor is h. Then the distance to the tumor is (r + h). Then, from the previous equations, we have

    [tex]\Delta P = \frac{-k}{2r^3}((r+h) - r) = \frac{-k}{2r^3}h[/tex]

    For there to be no difference in energy received by the skin and the tumor, this difference must be zero. There's no way that they can be zero unless k is zero (there is no radiation being emitted) or h is zero (the tumor is the same distance from the source as the skin).

    Now, when is this difference small and when is it large? What does this tell us about how far away we should set the source?

  6. Nov 16, 2004 #5
    if you are considering only inverse square losses in radiation intensity, then the relative intensity between the skin and tumour is
    where r is the source-tumour distance and h is the tumour depth (skin-tumour distance).

    as r increases, [tex]r-h \rightarrow r[/tex], so this ratio approaches unity.
    Last edited: Nov 16, 2004
  7. Nov 16, 2004 #6
    I think this tells us the ratio will decrease but the absolute value of the radiation difference will remain the same if the tumor is required to get the same amount of rads.
  8. Nov 16, 2004 #7
    then consider this.

    if r = source-tumour distance, h = tumour depth (skin-tumour distance), then the radiation intensities at the skin and the tumour are proportional to the inverse square of their distances.
    [tex]I_{tumour} \alpha \frac{1}{r^2}[/tex]
    [tex]I_{skin} \alpha \frac{1}{(r-h)^2}[/tex]
    The ratio of the two intensities approaches unity as r increases. The absolute difference between the two is
    [tex]I_{skin} - I_{tumour} \alpha \frac{1}{(r-h)^2} - \frac{1}{r^2}[/tex]
    as r increases, you again have [tex](r-h) \rightarrow r[/tex], so this difference approaches 0.

    there will still be a difference, but as your prof said, the differences become less important as r increases.

    plug in some numbers and see for yourself.
    Last edited: Nov 16, 2004
  9. Nov 16, 2004 #8
    What? The ratio of the values is different, but the values themselves are the same?

    2/5 = .4, and 5 - 2 = 3
    4/5 = .8, and 5 - 4 = 1
    5/5 = 1, and 5 - 5 = 0

    but maybe
    5/5 = .4, and 5 - 5 = 0.
    4/5 = .4, and 5 - 4 = 1.

    Plus, the equation doesn't tell you anything about the ratio. For a sufficiently smooth curve and a sufficiently small [itex]\Delta r[/itex], it approximates the difference quite well. You can even test it on functions you trust. Like 1/x^2. Or e^-x.

    Are you trying to get somebody to just tell you you're right, or are you trying to understand why you're wrong?

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