# Question 3a474

Sep 7, 2017

$\text{_39^88"Y}$

#### Explanation:

For starters, you know that you're dealing with a neutral atom, which means that the number of electrons that surround its nucleus must be equal to the number of protons present in its nucleus.

color(blue)(ul(color(black)("no. of protons = no. of electrons"))) " " -> " neutral atom"#

This means that your atom has

$\text{no. of protons} = 39$

As you know, the number of protons located inside the nucleus of an atom, i.e. its atomic number, gives you the identity of the element, so grab a Periodic Table and look for the element that has an atomic number, $Z$, equal to $39$.

You should be able to identify this element as yttrium, $\text{Y}$.

So, you know that you're dealing with an isotope of yttrium. In order to be able to figure out the identity of this isotope, you must calculate its mass number, $A$.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{mass number = no. of protons + no. of neutrons}}}}$

$A = 39 + 49 = 88$

To write this isotope in isotope notation, use

• the name of the element
• its atomic number
• the mass number of the isotope In your case, you have $\text{Y}$ as the chemical symbol for the element, $Z = 39$, and $A = 88$, so you will end up with

$\text{_39^88"Y" " " -> " " "yttrium-88}$