Answer
See the detailed answer below.
Work Step by Step
We are given a mysterious atom with four emission wavelengths:
- 310.0 nm
- 354.3 nm
- 826.7 nm
- 1240.0 nm
To determine the energy transitions from the given wavelengths (310.0 nm, 354.3 nm, 826.7 nm, and 1240.0 nm), we use the formula of
$$
E = \frac{hc}{\lambda}
$$
So we got the following:
- $ 310.0 \, \text{nm} $ corresponds to $ 4.00 \, \text{eV} $
- $ 354.3 \, \text{nm} $ corresponds to $ 3.50 \, \text{eV} $
- $ 826.7 \, \text{nm} $ corresponds to $ 1.50 \, \text{eV} $
- $ 1240.0 \, \text{nm} $ corresponds to $ 1.00 \, \text{eV} $
The task is to draw an energy level diagram where a ground $ p $ state (with $ l = 1 $) exists at 0 eV, an excited $ s $ state (with $ l = 0 $) at 3.50 eV, an excited $ d $ state (with $ l = 2 $) at 4.00 eV, and an excited $ p $ state at 5.00 eV.
Thus,
- The energy of the ground state is 0 eV.
- The $ s $ state has energy 3.50 eV.
- The $ d $ state has energy 4.00 eV.
- The excited $ p $ state has energy 5.00 eV.
The transitions between these states correspond to the four given wavelengths:
- $ 310.0 \, \text{nm} $ corresponds to an energy difference of 4.00 eV.
- $ 354.3 \, \text{nm} $ corresponds to an energy difference of 3.50 eV.
- $ 826.7 \, \text{nm} $ corresponds to an energy difference of 1.50 eV.
- $ 1240.0 \, \text{nm} $ corresponds to an energy difference of 1.00 eV.
Therefore,
$\bullet$ The excited $ p $ state at 5.00 eV can decay to:
1- The $ s $ state at 3.50 eV, releasing 1.50 eV (corresponding to 826.7 nm).
2- The $ d $ state at 4.00 eV, releasing 1.00 eV (corresponding to 1240 nm).
$\bullet$ The $ d $ state at 4.00 eV can decay only to the ground $ p $ state at 0 eV, releasing 4.00 eV (corresponding to 310.0 nm).
$\bullet$ The $ s $ state at 3.50 eV can also decay only to the ground $ p $ state at 0 eV, releasing 3.50 eV (corresponding to 354.3 nm).
See the figure below.
