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Minimum safe distance from a radioactive source

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    The potassium isotope 4219K has a half-life of 12 hr, and disintegrates with the emission of a γ-ray to form the calcium isotope 4220Ca. What other radiation besides γ-rays must be emitted? How many electrons, protons, and neutrons are there in an atom of the calcium isotope?

    The amount of radiation received in unit time by a person working near a radioactive source, commonly called the dose rate, is measured in rem hr-1. The safety regulations forbid dose rates in excess of 7.5 * 10-4 rem hr-1.The γ-ray dose rate from the 4219K source is found to be 3 * 10-3 rem hr-1 at a distance of 1 m. What is the minimum distance from this source at which it is safe to work?

    After how long will it be safe to work at a distance of 1 m from this source?

    Answers: 2.0 m; 24 hours.

    2. The attempt at a solution
    4219K → 4220Ca + 0-1β + 00γ.

    4220Ca: electrons = protons = 20, neutrons = 22.

    Minimum distance is (3 * 10-3) / (7.5 * 10-4) = 4. So 1 m from the source is 4 times more dangerous than it should be. So the distance should be increased 4 times, so the safe distance is 4 m. Why the answer is 2 m?

    I used A = A0 e- λ t to find time. The dose should decrease from 3 * 10-3 rem hr-1 to 7.5 * 10-4 rem hr-1. So: 7.5 * 10-4 = 3 * 10-3 e- (ln 2 / 12) t → t = 4.3 hours. Why not 24? I also calculated everything in seconds, not hours, still same result when I change the final answer to seconds.

    Why the distance is wrong and how to get the correct time?
    Last edited: Oct 31, 2016
  2. jcsd
  3. Oct 31, 2016 #2

    James R

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    It's a beta decay process, which produces ##\beta^-## particles (electrons), as you say. In the beta-minus decay, an anti-neutrino is also produced.
    The initial potassium isotope has 19 electrons. A beta particle (i.e. one more electron) is created in the decay process and is emitted from the system as ionising radiation. But the resulting calcium atom that is left over has only a changed nucleus and not a changed number of electrons surrounding that nucleus. So, the Calcium atom only has 19 electrons after the decay process.

    The intensity of radiation from a point source drops off with the inverse square of the distance from the source.

    The figure ##7.5\times 10^{-4}## doesn't appear in the problem statement that you've quoted. If it is correct, then your solution method is correct. However, I calculate the time as 24 hours, as required, using your numbers.
  4. Oct 31, 2016 #3


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    I don't see any definition for a safe dose rate. Is there information missing from the problem statement, or are there standard values from a table that you need to know?
  5. Oct 31, 2016 #4
    Very sorry, somehow I skipped an entire sentence while making the problem and then got distracted to check the text for mistakes,
  6. Oct 31, 2016 #5

    James R

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    A quick check:

    ##\frac{7.5\times 10^{-4}}{3.0\times 10^{-3}}=\frac{1}{4}## so we need a dose rate that is one-quarter of its initial value. After 1 half-life the rate will be half of what it was initially, and after 2 half-lives it will be 1/4, so we need 2 half lives, or 24 hours.
  7. Oct 31, 2016 #6
    Got it.

    4219K → 4220Ca + 0-1β + 00γ.

    4220Ca: electrons = 19 (since we need to look at K, which has 19 electrons), protons = 20, neutrons = 22.

    Not sure whether I understand this part. Is there a formula?

    Yes, I re-calculated and got 24 hours. Had some corrections in my notes so probably calculated it wrong : ).
  8. Nov 1, 2016 #7
    Could you please elaborate on this part? This is the last thing I don't understand quite well in this problem : ).
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