Question

If light of wavelength `"434 nm"` irradiates a hydrogen atom, what energy level transition will it make? What series is this in the emission line spectrum?

Answer 1

`n=5 -> n=2`

Explanation:

All you have to do here is to use the Rydberg formula for a hydrogen atom

`1/(lamda_"e") = R * (1/n_1^2 - 1/n_2^2)`

Here

• `lamda_"e"` is the wavelength of the emitted photon (in a vacuum)
• `R` is the Rydberg constant, equal to `1.097 * 10^(7)` `"m"^(-1)`
• `n_1` represents the principal quantum number of the orbital that is lower in energy
• `n_2` represents the principal quantum number of the orbital that is higher in energy

Notice that you need to have `n_1 < n_2` in order to avoid getting a negative value for the wavelength of the emitted photon.

So, convert the wavelength from nanometers to meters

`434 color(red)(cancel(color(black)("nm"))) * "1 m"/(10^9color(red)(cancel(color(black)("nm")))) = 4.34 * 10^(-7)` `"m"`

Rearrange the Rydberg formula to isolate the unknown variables

`1/n_1^2 - 1/n_2^2 = 1/lamda_"e" * 1/R`

Plug in your values to find

`1/n_1^2 - 1/n_2^2 = 1/(4.34 * color(blue)(cancel(color(black)(10^(-7))))color(red)(cancel(color(black)("m"))) * 1.097 * color(blue)(cancel(color(black)(10^7)))color(red)(cancel(color(black)("m"^(-1)))))`

`1/n_1^2 - 1/n_2^2 = 0.21`

This will be equivalent to

`(n_2^2 - n_1^2)/(n_1 * n_2)^2 = 21/100`

At this point, you have two equations with two unknowns, since

`{(n_2^2 - n_1^2 = 21), (n_1 * n_2 = sqrt(100)) :}`

Use the second equation to write

`n_1 = 10/n_2" "color(darkorange)("(*)")`

Plug this into the first equation to get

`n_2^2 - (10/n_2)^2 = 21`

`n_2^2 - 100/n_2^2 = 21`

This is equivalent to

`n_2^4 - 100 = 21 * n_2^2`

`n_2^4 - 21n_2^2 - 100 = 0`

If you take `n_2^2 = x`, you will have

`x^2 - 21x - 100 = 0`

Use the quadratic equation to get

`x_(1,2) = ( -(-21) +- sqrt( (-21)^2 - 4 * 1 * (-100)))/(2 * 1)`

`x_(1,2) = (21 +- sqrt(841))/2`

`x_(1,2) = (21 +- 29)/2 implies {( x_1 = (21 - 29)/2 = -4), (x_2 = (21 + 29)/2 = 25) :}`

Since we know that

`n_2^2 = x`

you can only use the positive solution here, so

`n_2^2 = 25 implies n_2 = 5`

According to equation `color(darkorange)("(*)")`, you will have

`n_1 = 10/5 = 2`

This means that your transition is taking place from

`n=5 -> n=2`

which is part of the Balmer series.