From Atkins-De Paula, Physical Chemistry, Ninth Edition, page 485

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Not sure what the books means as "percentage difference". Btw, this is my attempt

Fundamental vibration, in wavenumber, generic formula:

$\varepsilon_{\nu=1}-\varepsilon_{\nu=0} = \bar{\omega}$

$\bar{\omega} = \dfrac{1}{2\pi c} \sqrt{\dfrac{k}{\mu}}$

$\text{let }1) \ce{^{23}Na}\ce{^{35}Cl} \,\,\, \mathrm{and} \,\,\, 2) \ce{^{23}Na}\ce{^{37}Cl}$

$\mu_1 = 2.31 \times 10^{-26} \, \mathrm{kg}$

$\mu_2 = 2.37 \times 10^{-26} \, \mathrm{kg}$

$\text{percentage difference} = 100 - \dfrac{\omega_2}{\omega_1} = 100 - \dfrac{\sqrt{1/\mu_2}}{\sqrt{1/\mu_1}}$

According to what is reported in the book, the solution is 1.089 percent. Is this procedure right?

  • $\begingroup$ Percentage difference is the ordinary relative difference, just expressed in percentage. The only problem is what to take as the reference 100%? Value A, value B or their average ? // BTW if both omegas are the same the percentage difference is hardly 99% as it is 0%. $\endgroup$
    – Poutnik
    Commented Jul 23, 2023 at 12:55
  • $\begingroup$ 4 is by 20% smaller (percentage difference) than 5. 5 is by 25% bigger than 4. They are respectively (100/9)% smaller/bigger than 4.5 as the average of 4 and 5. $\endgroup$
    – Poutnik
    Commented Jul 23, 2023 at 13:01
  • $\begingroup$ I formatted the text as you requested. Now, is my reasoning right? $\endgroup$ Commented Jul 23, 2023 at 13:07
  • $\begingroup$ Read again the second part of my second comment. $\endgroup$
    – Poutnik
    Commented Jul 23, 2023 at 13:09
  • $\begingroup$ By the last equation line you say -- by other words -- that 5 is by 99% bigger/smaller than 5. That is not true, is it? $\endgroup$
    – Poutnik
    Commented Jul 23, 2023 at 13:17

1 Answer 1


The relative percentage difference (RPD) of $b$ with respect to $a$, where the value $a$ is given or taken as the reference 100% value(typically if $b$ is deviation from $a$), is:

$$\text{RPD} = \frac{b-a}{a} \cdot 100 \%.$$

In your case:

$$\text{RPD} = \frac{\bar{\omega}_ 2-\bar{\omega}_ 1}{\bar{\omega}_ 1} \cdot 100 \%.$$

In case the reference 100% value is not given, is unclear, or is explicitly said so, as the reference can be taken the arithmetic average of both values(typically if both have equal, parallel status). Then:

$$\text{RPD} = \frac{2(b-a)}{a+b} \cdot 100 \%$$

In your case:

$$\text{RPD} = \frac{2(\bar{\omega}_ 2-\bar{\omega}_ 1)}{\bar{\omega}_ 1+\bar{\omega}_ 2} \cdot 100 \%.$$

Regarding how exactly it was meant by the book, I just refer to one of my canned comments:

A asking B "What does C mean by X?" may not provide a satisfactory nor reliable answer. A should ask C.

The hint can be which of 3 possible options for the reference fits the given result the best.


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