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UnkleRhaukus

Does beta decay violate Kirchhoff's current law?

  • one year ago
  • one year ago

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  1. rajathsbhat
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    what makes you think like that?

    • one year ago
  2. UnkleRhaukus
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    electrons (or positrons) , and hence current is flowing out of the region

    • one year ago
  3. shivam_bhalla
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    @UnkleRhaukus, didn't get you. a neutral atom which on decay emits electrons/positrons, hence resulting in current. Now how is kirchoff's law violated in your opinion?

    • one year ago
  4. PaxPolaris
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    Kirchhoff's laws apply only to \(\large \text {electrical circuits}\).

    • one year ago
  5. yakeyglee
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    It's not a violation, because the decay itself is a current. Remeber the continuity equation, which is closely tied to the conservation of charge: \[\vec \nabla \cdot \vec J = -\frac{\partial \rho}{\partial t}\]

    • one year ago
  6. PaxPolaris
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    I think he means that current flowing into the decaying atom(0) is not the same as current flowing out.. but circuit laws shouldn't apply to the atom

    • one year ago
  7. yakeyglee
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    It's honestly more of a conservation of charge concept. Think about a capacitor. Even though current doesn't flow through, charge is still conserved as it collects on the plates, and thus we don't think twice about it when we apply Kirchoff's Current Law. If we think of each of the plates themselves as junctions, we find that the current flowing in is NOT the current flowing out. It's a similar concept here. It's not that it's a violation, it's rather that Kirchoff's Current Law has no place in microscopic electrodynamics.

    • one year ago
  8. Carl_Pham
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    How can it? Charge is always conserved. In beta decay you emit an electron, and the nucleus acquires another positive charge. That either leaves you with a cation -- in which case you have a current of both electrons and positive anions, i.e. a neutral current -- or the cation absorbs a nearby electron, which cancels out the charge of the beta particle. It's true when you think about nuclear physics, you tend to ignore the behaviour of any orbital electrons, or indeed any electrons at thermal (and not nuclear) energies. But they're there, and in cases interact to balance the books.

    • one year ago
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