Question

11.49 A typical rotational transition wavenumber is on the order of 1 cm$^{-1}$ and a typical vibrational transition wavenumber is on the order of 1000 cm$^{-1}$. Calculate the energy (in kJ mol$^{-1}$) for typical rotational and vibrational transitions. Compare the period of rotation (the time required for one revolution) with the period of vibration.

          11.49 A typical rotational transition wavenumber is on the order of 1 cm$^{-1}$ and a typical vibrational transition wavenumber is on the order of 1000 cm$^{-1}$. Calculate the energy (in kJ mol$^{-1}$) for typical rotational and vibrational transitions. Compare the period of rotation (the time required for one revolution) with the period of vibration.

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11.49 A typical rotational transition wavenumber is on the order of 1 cm^-1 and a typical vibrational transition wavenumber is on the order of 1000 cm^-1. Calculate the energy (in kJ mol^-1) for typical rotational and vibrational transitions. Compare the period of rotation (the time required for one revolution) with the period of vibration.

Added by Tyler P.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Suman K

06/27/2022

Step 1

626 \times 10^{-34} \, \text{J s}\) (Planck's constant), \(c = 3 \times 10^8 \, \text{m/s}\) (speed of light), and \(\tilde{\nu} = 10^2 \, \text{m}^{-1}\) (wave number for rotational transition). \[E_{\text{rotational}} = 6.626 \times 10^{-34} \times 3 \times 10^8 Show more…

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Transcript

00:01 Hello students in the question there are three parts.

00:03 In first and second part we have to calculate the energy for rotational transition and vibrational transition.

00:11 And in the third part we have to compare the time period for them.

00:17 So let's get started.

00:19 We have one formula that is t equals to 2 pi divided by c multiplied by the wave number.

00:29 So this is the formula.

00:31 Here t is period of rotation or vibration.

00:38 Period of rotation or vibration.

00:51 Now c is the speed of the wave.

00:56 Speed of wave.

01:00 This is equal to 3 multiplied by 10 to the power 8 meter per second.

01:06 Now this is the wave number.

01:13 Also in the equation it is given the wave number for rotational transition is 1 cm inverse.

01:24 That is, if we convert it into meters, it will be 10 square meter inverse.

01:33 Now, we have also the wave number for vibrational transition.

01:41 It will be equal to, if we convert centimeter into meter, it will be equal to 10 to the power 5 meter inverse.

01:52 Now let's calculate the energy.

01:59 So, first we'll calculate the energy.

02:01 Energy for rotational transition.

02:04 Energy for rotational transition.

02:07 The formula we have is nahhc multiplied by the wave number.

02:13 Then the answer will come into the unit jule per mole.

02:19 Now we'll put the values 6 .626 multiplied by 10 to the power minus 34.

02:31 This is the value of n...

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