Question

How many different photon energies would emerge from a collection of hydrogen atoms that occupy the lowest four energy states $\mathbf{(}\mathbf{N}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{4}\mathbf{)}$ ? (You need not calculate the energies of each states.

Short answer

The six different photon energies would emerge from a collection of hydrogen atoms that occupy the lowest four energy states$(\mathrm{N}=1,2,3,4)$ .

Step-by-step solution

Understanding the concept

If the hydrogen atom gets drop from higher energy state to lower then it will emit a photon with energy equal to ${\mathbf{E}}_{\mathbf{f}}\mathbf{-}{\mathbf{E}}_{\mathbf{i}}$.

Here ${\mathbf{E}}_{\mathbf{f}}$is the energy of g hydrogen atom at higher level and ${\mathbf{E}}_{\mathbf{i}}$ is the energy of hydrogen atom at lower level.

Calculate the number of different photon energies.

Draw the energy transition diagram of hydrogen atoms

From the above diagram it is clear that the six different photon energies, corresponding to transitions$4\to 3$ , $4\to 2$,$4\to 1$ ,$3\to 2$ ,$3\to 1$ and$2\to 1$ are emerged.

Therefore six different photon energies would emerge from a collection of hydrogen atoms that occupy the lowest four energy states $(\mathrm{N}=1,2,3,4)$.