# What the book means for "percentage difference"?

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

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?

• 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%. Commented Jul 23, 2023 at 12:55
• 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. Commented Jul 23, 2023 at 13:01
• I formatted the text as you requested. Now, is my reasoning right? Commented Jul 23, 2023 at 13:07
• Read again the second part of my second comment. Commented Jul 23, 2023 at 13:09
• 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? Commented Jul 23, 2023 at 13:17

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:

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

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

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:

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

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