Question

Question #3a474

Answer 1

`""_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

`"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, `"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`.

`color(blue)(ul(color(black)("mass number = no. of protons + no. of neutrons")))`

In your case, you have

`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 `"Y"` as the chemical symbol for the element, `Z = 39`, and `A = 88`, so you will end up with

`""_39^88"Y" " " -> " " "yttrium-88"`