Question #9f672

1 Answer
Stefan V.
Jun 20, 2017

Here's what I got.

Explanation:

The thing to keep in mind here is that bismuth-203 undergoes beta positive decay, or positron emission, not beta minus decay, which more often than not is simply called beta decay.

https://en.wikipedia.org/wiki/Isotopes_of_bismuth

When a radioactive nuclide undergoes positron emission, a proton is converted to a neutron and a positron, e+, the antiparticle of an electron, and an electron neutrino, νe, are emitted from the nucleus.

![https://cnx.org/contents/947b85ec-f1d9-40b9-8b81-df05d613440c@3/31-5-nuclear-decay-and-conservation-laws](useruploads.socratic.org)

This means that--keep in mind that charge and mass are conserved in a nuclear reaction!

  • the atomic number of the nuclide, Z, will decrease by 1
  • the mass number of the nuclide, A, will remain unchanged

So, you can write

203183Bi→AZ?+01e+νe

The atomic number decreases by 1, so

83=Z+1⇒Z=82

The mass number remains unchanged, so

203=A+0⇒A=203

Grab a Periodic Table and look for the element with the atomic number equal to 82. You'll find that you're dealing with lead, Pb. This implies that the resulting isotope is lead_203.

The balanced nuclear equation that describes the position emission of bismuth-203 will thus look like this

203183Bi→203182Pb+01e+νe

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