Question

The ionization energy of an atom is known to be $5.5eV$. The emission spectrum of this atom contains only the four wavelengths $310.0nm,354.3nm,826.7nm$and $1240.0nm$. Draw an energy-level diagram with the fewest possible energy levels that agrees with these experimental data. Label each level with an appropriate $l$quantum number. Hint: Don’t forget about the $\overrightarrow{l}$selection rule.

Short answer

$E_1-E_2=4.006eV,3.5eV,1.5eV,1eV$

Step-by-step solution

Wavelength of photon emitted:

The allowed transitions are determined form the following usual selection rules

$$\begin{gathered} ∆L=\pm 1 \\ ∆S=0 \\ ∆J=0,\pm 1 \end{gathered}$$

If the above selection rules are not satisfied, no other transitions are allowed between the energy levels.

The wavelength of photon emitted as,

$\lambda =\frac{hc}{E_2-E_1}$

Rearranging the above equation for the energy difference,

$E_2-E_1=\frac{hc}{\lambda }$.....(1)

Here, $h$is planck's constant $(4.14\times {10}^{-15}eV.s)$and $c$is speed of light in vacuum$\left(3\times {10}^{8}m/s\right)$.

Conversion of unit wavelength:

Convert the unit of wavelength from $nm$to $m$.

role="math" $$\begin{gathered} \lambda =310nm \\ =310nm\left({10}^{-9}m/nm\right) \\ =310\times {10}^{-9}nm \end{gathered}$$

Substitute $310\times {10}^{-9}nm$for role="math" localid="1648626935471" $\lambda ,414\times {10}^{-15}eV.s$for $\lambda$and $3\times {10}^{8}m/s$for $c$in the above equation.

role="math" localid="1648627016505" $$\begin{gathered} E_1-E_2=\frac{\left(4.14\times {10}^{-15}eV.s\right)\left(3\times {10}^{8}m/s\right)}{\left(310\times {10}^{-9}nm\right)} \\ =4.006eV \end{gathered}$$

Conversion of unit wavelength:

Convert the unit of wavelength from $nm$to $m$.

$$\begin{gathered} \lambda =354.3nm \\ =354.3nm\left({10}^{-9}m/nm\right) \\ =354.3\times {10}^{-9}nm \end{gathered}$$

Substitute $354.3\times {10}^{-9}nm$for role="math" localid="1648627243412" $\lambda ,414\times {10}^{-15}eV.s$for $\lambda$and $3\times {10}^{8}m/s$for $c$in the above equation.

$$\begin{gathered} E_1-E_2=\frac{\left(4.14\times {10}^{-15}eV.s\right)\left(3\times {10}^{8}m/s\right)}{\left(354\times {10}^{-9}nm\right)} \\ =3.5eV \end{gathered}$$

Conversion of unit wavelength:

Convert the unit of wavelength from $nm$to $m$.

localid="1648628463405" $$\begin{gathered} \lambda =826.7nm \\ =826.7nm\left({10}^{-9}m/nm\right) \\ =826.7\times {10}^{-9}nm \end{gathered}$$

Substitute $826.7\times {10}^{-9}m$for localid="1648628524608" $\lambda ,414\times {10}^{-15}eV.s$for $\lambda$and $3\times {10}^{8}m/s$for $c$in the above equation.

localid="1648628584457" $$\begin{gathered} E_1-E_2=\frac{\left(4.14\times {10}^{-15}eV.s\right)\left(3\times {10}^{8}m/s\right)}{\left(826.7\times {10}^{-9}nm\right)} \\ =1.5eV \end{gathered}$$

Conversion of unit wavelength:

Convert the unit of wavelength from $nm$to $m$.

$$\begin{gathered} \lambda =1240nm \\ =1240nm\left({10}^{-9}m/nm\right) \\ =1240\times {10}^{-9}nm \end{gathered}$$

Substitute $1240\times {10}^{-9}m$for $\lambda ,414\times {10}^{-15}eV.s$for $\lambda$and $3\times {10}^{8}m/s$for $c$in the above equation.

$$\begin{gathered} E_1-E_2=\frac{\left(4.14\times {10}^{-15}eV.s\right)\left(3\times {10}^{8}m/s\right)}{\left(1240\times {10}^{-9}m\right)} \\ =1eV \end{gathered}$$

Definition with graph explanation:

For the wavelength $310nm$, the difference in energy levels is $4eV$so we can think of it as the maximum transition. So if we take the ground state as $0eV$. The one state will be of $4eV$.

There is possibility of another state having energy $3,5eV$.If we take the energy level as $2.5eV$.

Then we can get the exact energy levels corresponding to given data as

$$\begin{gathered} \left(1\right)4eV\to 04eV-0=4eV \\ \left(2\right)3.5eV\to 03,5eV-0=3.5eV \\ \left(3\right)4eV\to 2.5eV4eV-2.5eV=1.5eV \\ \left(4\right)3.5eV\to 2.5eV3.5eV-2.5eV-1eV \end{gathered}$$

So we can level the following energy levels:

Labeling each energy:

Let us label each energy level with appropriate $l$. For this we need to consider $∆l=0$.

Let $0eV$state is of data-custom-editor="chemistry" $\mathrm{l}=1$l.e., $p$State can be $l=2$i.e. $d$. Then $2.5eV$state should also be of $l=1$i.e., $P$.

Diagram:

So we can have the following diagram: