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The mass spectrum in Bromine, with the molecules $\ce{^{158}Br2+}$, $\ce{^{160}Br2+}$ and $\ce{^{162}Br2+}$:

Bromine Mass Spectrum 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?

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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.

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  • $\begingroup$ Because there are only ever two isotopes of Bromine, the outcome of 160 amu is currently illustrated as more likely than 162 or 158. $\endgroup$
    – Moobius Strip
    Sep 7, 2020 at 15:50
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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.

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  • $\begingroup$ @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. $\endgroup$
    – Joe
    Nov 21, 2017 at 20:25
  • $\begingroup$ I see what you did, well OK, I just wanted to say that this question shouldn't be here at all. $\endgroup$
    – Mithoron
    Nov 21, 2017 at 20:46
  • $\begingroup$ 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". $\endgroup$
    – Chris
    Nov 22, 2017 at 11:46

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