Question

Question #c7335

Answer 1

For `n`=2 to `n`=1 `->` `lambda` = 121 nanometer

Explanation:

The Rydberg equation applies for this problem:

`DeltaE = -R_H * (1/n_f^2 - 1/n_i^2)`

where:
`R_H` = `2.18 * 10^(-18)J`
`n_f` = the final energy level
`n_i` = the initial energy level

So for `n`=2 to `n`=1:

`DeltaE = -2.18*10^-18 J * (1/1 - 1/4)`

`DeltaE = -1.65*10^-18 J`

We know that `DeltaE` is for one photon. The value is negative because energy is being released.

And by the:

`E=(hc)/(lambda)`

`lambda=(6.626*10^-34*J*s*2.99*10^8*m*s^-1)/(1.65*10^-18 J)`

`lambda=1.21*10^-7*m`

We know the visible part of the spectrum can clear ranges between 300-700 nanometer, nm.

`1.21*10^-7m*(10^9nm)/(1m)=120.5nm`

The transition does not emit a wavelength of light which is in the visible range. A wavelength of `121` nanometer approx is in the ultraviolet range of the spectrum.

As you can see these problem are not hard, but the work is tedious. From looking at my work in solving n=2 to n=1, I hope that you can solve the others using the exact information and equation which I have provided.

Remember that Rydberg equation only apply to the hydrogen atom.