91Ó°ÊÓ

An electron can go from one energy level to another, but it cannot move back and forth between them.

We can begin the solution by expressing the photon's energy.

${\mathrm{E}}_{\mathrm{ph}}=\frac{\mathrm{hc}}{\mathrm{λ}}$

We derive the following energies by substituting given wavelengths$2.48\mathrm{eV},4.14\mathrm{eV},6.20\mathrm{eV}$ which are depicted in the graph

The spectrum of frequencies of electromagnetic radiation released by an electron transitioning from a high energy state to a lower energy state is known as the emission spectrum of a chemical element or chemical compound.

The following expression can be used to calculate the wavelengths seen in the atom's emission spectrum:

$$\begin{gathered} {\mathrm{λ}}_{\mathrm{nm}}=\frac{\mathrm{hc}}{{\mathrm{E}}_{\mathrm{n}}−{\mathrm{E}}_{\mathrm{m}}} \\ {\mathrm{λ}}_{21}=\frac{\mathrm{hc}}{{\mathrm{E}}_2−{\mathrm{E}}_1}(\text{emission}2→1) \\ {\mathrm{λ}}_{21}=\frac{\left(6.63×{10}^{−34}\right)×\left(3×{10}^8\right)}{\left(2.49\right)×\left(1.6×{10}^{−19}\right)−0}\text{(substitute)} \\ {\mathrm{λ}}_{21}=500\mathrm{nm} \\ {\mathrm{λ}}_{31}=\frac{\mathrm{hc}}{{\mathrm{E}}_3−{\mathrm{E}}_1}(\text{emission}3→1) \\ {\mathrm{λ}}_{31}=\frac{6.63â‹\dots {10}^{−34}â‹\dots 3â‹\dots {10}^8}{4.14â‹\dots 1.6â‹\dots {10}^{−19}−0}\text{(substitute)} \\ {\mathrm{λ}}_{31}=300\mathrm{nm} \\ {\mathrm{λ}}_{41}=\frac{\mathrm{hc}}{{\mathrm{E}}_4−{\mathrm{E}}_1}(\text{emission4}→1) \\ {\mathrm{λ}}_{41}=\frac{6.63â‹\dots {10}^{−34}â‹\dots 3â‹\dots {10}^8}{6.21â‹\dots 1.6â‹\dots {10}^{−19}−0}\left(\mathrm{substitute}\right) \\ {\mathrm{λ}}_{41}=200\mathrm{nm} \\ {\mathrm{λ}}_{32}=\frac{\mathrm{hc}}{{\mathrm{E}}_3−{\mathrm{E}}_2}(\text{emission3}→2) \\ {\mathrm{λ}}_{32}=\frac{6.63â‹\dots {10}^{−34}â‹\dots 3â‹\dots {10}^8}{4.14â‹\dots 1.6â‹\dots {10}^{−19}−2.49â‹\dots 1.6â‹\dots {10}^{−19}}\left(\mathrm{substitute}\right) \\ {\mathrm{λ}}_{32}=752\mathrm{nm} \\ {\mathrm{λ}}_{42}=\frac{\mathrm{hc}}{{\mathrm{E}}_4−{\mathrm{E}}_2}(\text{emission}4→2) \\ {\mathrm{λ}}_{42}=\frac{6.63â‹\dots {10}^{−34}â‹\dots 3â‹\dots {10}^8}{6.21â‹\dots 1.6â‹\dots {10}^{−19}−2.49â‹\dots 1.6â‹\dots {10}^{−19}}\text{(substitute)} \\ {\mathrm{λ}}_{42}=334\mathrm{nm} \\ {\mathrm{λ}}_{43}=\frac{\mathrm{hc}}{{\mathrm{E}}_4−{\mathrm{E}}_3}(\text{emission}4→3) \\ {\mathrm{λ}}_{43}=\frac{6.63â‹\dots {10}^{−34}â‹\dots 3â‹\dots {10}^8}{6.21â‹\dots 1.6â‹\dots {10}^{−19}−4.14â‹\dots 1.6â‹\dots {10}^{−19}}\text{(substitute)} \\ {\mathrm{λ}}_{43}=599\mathrm{nm} \end{gathered}$$