Question

"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: εν=1−εν=0=ˉω ˉω=12πc√kμ μ1=2.31×10−26kg μ2=2.37×10−26kg percentage difference=100−ω2 ω1=100−√1/μ2 According to what is reported in the book, the solution is 1.089 percent. Is this procedure right?

Answer 1

The percentage difference between fundamental vibrations is calculated using the difference in vibrational frequencies for two different reduced masses, divided by the frequency corresponding to one of the masses and expressed as a percentage.

Step-by-step explanation:

The concept of "percentage difference" in this context refers to the relative difference in the vibrational frequency (in wavenumbers) of a molecule due to a change in its reduced mass (μ).

To calculate the percentage difference, you first calculate vibrational frequency for each reduced mass using the formula ν = √(k/μ) / (2πc), where k is the force constant, c is the speed of light, and μ is the reduced mass. Then, you find the difference between the two frequencies. Lastly, you divide this difference by one of the frequencies and multiply by 100 to express it as a percentage.

Following your procedure, the percentage difference between the fundamental vibrations for two different reduced masses μ1 and μ2 is calculated incorrectly. The correct formula to find the percentage difference is 100 * (| ν1 - ν2 | / ν1), where ν1 and ν2 are the vibrational frequencies for μ1 and μ2, respectively.

If we assume the force constant k remains the same for both cases, the vibrational frequency is inversely proportional to the square root of the reduced mass. Therefore, the actual calculation should take the difference between the inverse of the square roots of the reduced masses, not just one of them as in the original attempt.