Question #aa1e9

1 Answer
Jul 2, 2016

Only diagram (3) can represent the emission spectrum shown.


A hydrogen atom emits energy when the electron drops from a higher energy level to a lower energy level.

That rules out diagram (5), because here the electron is absorbing energy and jumping to higher energy levels.

Diagram (5) represents an absorption spectrum.

An important energy formula is

#color(blue)(|bar(ul(color(white)(a/a) E = hf color(white)(a/a)|)))" "#


#E# is the energy, #h# is Planck's constant, and #f# is the frequency of the light.

Another important formula is

#color(blue)(|bar(ul(color(white)(a/a) fλ = c color(white)(a/a)|)))" "#

where #λ# is the wavelength of the light and #c# is the speed of light.

We can rearrange this formula to get

#f = c/λ#

If we insert this into the first equation, we get a third formula:

#color(blue)(|bar(ul(color(white)(a/a) E = (hc)/λ color(white)(a/a)|)))" "#

This shows that the energy is inversely proportional to the wavelength of the light.

Thus, a longer wavelength corresponds to a smaller energy.

Peak (d) has the longest wavelength, so it corresponds to the smallest energy difference.

This rules out diagrams (2) and (4), because peak (d) in these diagrams corresponds to the largest energy difference.

The only possibility is diagram (3), because, in it, peak (d) corresponds to the smallest energy difference.