# Question #17b01

May 15, 2017

Here's what I got.

#### Explanation:

You're dealing with the beta minus decay of neptunium-238.

In beta minus decay, a neutron located inside the nucleus of a radioactive nuclide is being converted to a proton. The process releases an electron, also called a beta particle, and an electron antineutrino, ${\overline{\nu}}_{e}$.

$\textcolor{w h i t e}{a}$

So, you know that you're starting with

$\text{_ (color(white)(1)color(darkgreen)(93))^color(blue)(238)"Np" -> ""_ color(darkgreen)(Z)^color(blue)(A)"?" + ""_ color(darkgreen)(-1)^(color(white)(-)color(blue)(0))"e} + {\overline{\nu}}_{e}$

Your goal here is to figure out the values of $A$ and $Z$, the mass number and the atomic number, respectively, of the resulting nuclide.

As you know, mass and charge must be conserved in nuclear equations. This means that you can write

$\textcolor{\mathrm{da} r k g r e e n}{93 = Z + \left(- 1\right)} \to$ conservation of charge

$\textcolor{b l u e}{238 = A + 0} \textcolor{w h i t e}{a a a} \to$ conservation of mass

You can thus say that

$Z = 93 + 1 = 94 \text{ }$ and $\text{ } A = 238$

Grab a Periodic Table and look for the element that follows neptunium. You'll find that this element is plutonium, $\text{Pu}$. The resulting nuclide is plutonium-238.

You can now complete the nuclear equation that describes the beta minus decay of neptunium-238

$\text{_ (color(white)(1)color(darkgreen)(93))^color(blue)(238)"Np" -> ""_ (color(white)(1)color(darkgreen)(94))^color(blue)(238)"Pu" + ""_ color(darkgreen)(-1)^(color(white)(-)color(blue)(0))"e} + {\overline{\nu}}_{e}$