# Question #aa1e9

##### 1 Answer

#### Answer:

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

#### Explanation:

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)|)))" "#

where

Another important formula is

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

where

We can rearrange this formula to get

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.