# Question #9f672

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, ${\text{e}}^{+}$, the antiparticle of an electron, and an electron neutrino, ${\nu}_{\text{e}}$, are emitted from the nucleus.

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

$\text{_ (color(white)(1)83)^203"Bi" -> ""_ Z^A"?" + ""_ 1^0"e" + nu_ "e}$

The atomic number decreases by $1$, so

$83 = Z + 1 \implies Z = 82$

The mass number remains unchanged, so

$203 = A + 0 \implies A = 203$

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, $\text{Pb}$. 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

$\text{_ (color(white)(1)83)^203"Bi" -> ""_ (color(white)(1)82)^203"Pb" + ""_ 1^0"e" + nu_ "e}$