# Question #b1ea6

Apr 17, 2016

Here's what I got.

#### Explanation:

Isotope notation allows you to identify various isotopes by using atomic number, $Z$, chemical symbol, and mass number, $A$. So, let's start with antimony-125. Grab a periodic table and look for antimony, $\text{Sn}$. You'll find it located in period 5, group 14.

Antimony has an atomic number equal to $50$, which means that any atom that contains $50$ protons in its nucleus will be an atom of antimony.

Since you know that the given isotope has a mass number equal to $125$, you can say that its isotope notation will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{_(color(white)(a)50)^125"Sn}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ antimony-125

Next up, a bismuth nucleus that contains $120$ neutrons. Look for bismuth, $\text{Bi}$, in the periodic table. You'll find it located in period 6, group 16. Bismuth has an atomic number of $83$, which tells you that its nucleus contains $83$ protons.

No,w to get he mass number of this isotope, add the number of protons and the number of neutrons located in the nucleus

$A = Z + N$

$A = 83 + 120 = 203$

The isotope notation for this particular isotope will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{_(color(white)(a)83)^203"Bi}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ bismuth-203

Finally, a nucleus of element 35 that contains $46$ neutrons. Once again, pick the a periodic table and look for the element that has an atomic number equal to $35$.

You'll find that this element is bromine, $\text{Br}$, located in period 4, group 17.

Calculate the isotope's mass number by using the number of protons and the number of neutrons it contains in its nucleus

$A = 35 + 46 = 81$

The isotope notation for this isotope will be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{_ 35^81"Br}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ bromine-81