Question

5-18. The following data are obtained from the infrared spectrum of 1271CI. Using the method of Problem 5-17, determine the values ofae and, from these data. Transition Frequency/cm1 381.20 759.60 1135.00 1507.40 1877.00 0?4

Answer 1

Alright starting with 5.17, the energy required for the molecule jump any vibrational level from v = 0, to v = 1,2,3,...

$\tilde{\omega}_{obs} = G(v) - G(0) = \tilde{\omega}_ev - \tilde{\chi}_e\tilde{\omega}_ev(v+1)$

Dividing the equation with on both sides,

$\frac{\tilde{\omega}_{obs}}{v} = \tilde{\omega}_e - \tilde{\chi}_e\tilde{\omega}_e(v+1)$

Which can be compared to the equation of a line, with slope and intercept.

This gives us the value of what we need very conveniently,

$\tilde{\chi}_e\tilde{\omega}_e = -m$
$\tilde{\omega}_e = c$

It is apparent that since it is anharmonic oscillator, as the wave-number rises, the energy gaps between two levels starts decreasing up to a point where a continuum is obtained.

(5.18) Let us first write the data we need,

Data we need

	$\tilde{\omega}_{obs}/v$
2	381.20
3	379.80
4	378.33
5	376.75
6	375.40

The graph I've plotted is a crude one on some online plotting website. A better job can be there, but here will work just fine.

Calculating slope should be easy since the line quite linear,

Click to enter Plot title 384 382 380 378 376 374 372 V+1

$m = -\frac{381.20-375.4}{6-2} = -1.45$

$\therefore -m = \tilde{\chi}_e\tilde{\omega}_e = 1.45 \ cm^{-1}$

The intercept can be extrapolated as,

$c \sim 384\ cm^{-1} = \tilde{\omega}_e$