Question

First vibrational level for NaH lies at 1.154 × 10-20 J and the second vibrational level lies at 3.406 × 10-20 J. What is the frequency of the photon emitted when a molecule of NaH drops from the second vibrational level to the first vibrational level?

Answer 1

The frequency of the `NaH` molecule will be `3.340 * 10^(13)"s"^(-1)`.

When dropping from the second vibrational level to the first, energy will be released, since the second vibrational level is higher in energy than the first.

The difference in energy between the two vibrational levels will be equal to the energy of the emitted photon. So,

`DeltaE = E_2 - E_1`

`DeltaE = 3.406 * 10^(-20)"J" - 1.154 * 10^(-20)"J" = 2.252 * 10^(-20)"J"`

By definition, a photon's energy is proportional to its frequency according to the equation

`E = h * nu`, where

`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`;
`nu` - the frequency of the photon.

So, plug your values into the equation and solve for `nu`

`E = h * nu => nu = E/h`

`nu = (2.252 * 10^(-20)cancel("J"))/(6.626 * 10^(-34)cancel("J") * "s") = color(green)(3.340 * 10^(13)"s"^(-1))`

If you want, you can express the frequency in Hz, which means that you'll have

`nu = "33.40 THz"` `->` 33.40 terahertz