# Why is the probability for the signal at 160 in the bromine mass spectrum twice as high as for the signals at 158 and 162?

The mass spectrum in Bromine, with the molecules $\ce{^{158}Br2+}$, $\ce{^{160}Br2+}$ and $\ce{^{162}Br2+}$: As you can see, the $\ce{^{160}Br2+}$ is almost double in intensity compared to the $\ce{^{158}Br2+}$ and the $\ce{^{162}Br2+}$ peak.

The book I am reading simply states that this is because

The probability of two different isotopes occurring in a $\ce{Br2}$ molecule are twice that of the same isotope appearing in a $\ce{Br2}$ molecule.

This is supported by the $\ce{^{160}Br2+}$ peak, formed from the $\ce{^{79}Br}$ and $\ce{^{81}Br}$ isotopes. Likewise, $\ce{^{158}Br2+}$ peak is formed from two $\ce{^{79}Br}$ isotopes and $\ce{^{162}Br2+}$ is formed from two $\ce{^{81}Br}$ isotopes.

However, I am confused by the explanation given by the book above. Why is the probability of two different isotopes occurring in a $\ce{Br2}$ molecule twice that of the same isotope appearing in a $\ce{Br2}$ molecule?

All possible arrangements of $$\ce{Br2}$$ molecule:

• $$\displaystyle 79 + 79 = 158$$
• $$\displaystyle \color{red}{79 + 81} = 160$$
• $$\displaystyle \color{red}{81 + 79} = 160$$
• $$\displaystyle 81 + 81 = 162$$

The amount of $$\ce{^{79}Br}$$ and $$\ce{^{81}Br}$$ in nature is roughly the same, thus each permutation is equally probable. There are two arrangements that lead to $$160$$. While $$158$$ and $$162$$ each have only one arrangement. Therefore $$160$$ is twice as likely to be found compared to other masses.

• Because there are only ever two isotopes of Bromine, the outcome of 160 amu is currently illustrated as more likely than 162 or 158. – Frogbert Sep 7 at 15:50

A way to understand this that may be familiar is that of the Punnett square from biology, since the two isotopes have nearly 50/50 split in nature.

\begin{array}{c|cc} & \ce{^{79}Br} & \ce{^{81}Br} \\\hline \ce{^{79}Br} & \ce{^{158}Br} & \ce{^{160}Br} \\ \ce{^{81}Br} & \ce{^{160}Br} & \ce{^{162}Br} \\ \end{array}

When breeding two hybrids (Aa x Aa), it's twice as likely to get a hybrid (Aa) than to get either homozygote. Similarly, here you have twice the chance to get a 'hybrid' $\ce{^{160}Br}$ than a particular 'homozygote' $\ce{^{158}Br}$ or $\ce{^{162}Br}$ .

However, I would disagree with the wording of the statement:

The probability of two different isotopes occurring in a $\ce{Br2}$ molecule are twice that of the same isotope appearing in a $\ce{Br2}$ molecule.

The probability is actually identical of two different isotopes occurring and any pair of identical isotopes occurring. This could be worded better:

The probability of two different isotopes occurring in a $\ce{Br2}$ molecule are twice that of a particular same isotope appearing in a $\ce{Br2}$ molecule.

• @Mithoron I just thought it might be a good (different) way to visualize it for someone who might be used to seeing this from high school biology, but not used to it in this context. – Joe Nov 21 '17 at 20:25
• I see what you did, well OK, I just wanted to say that this question shouldn't be here at all. – Mithoron Nov 21 '17 at 20:46
• Your better wording uses some not great english: "of a particular same isotope" just doesn't read well. Better might be "twice that of a particular isotope appearing twice in a $\ce{Br2}$ molecule". – Chris Nov 22 '17 at 11:46