The brilliant color of fireworks is produced by shells containing lithium carbonate (Li2CO3) that are
launched into the night sky and explode. The explosions produce enough energy to vaporize the Li2CO3
and produce excited state lithium atoms. These atoms quickly lose energy as they transition from their
excited states to their ground states, emitting photons of red light that have a wavelength of 670.8 nm.
a) (3 marks) Calculate the energy (Joules) in each photon of red light emitted by an excited lithium
atom.
b) (2 marks) Calculate the energy of one mole of these photons (kJ/mol).
c) (1 mark) Li atoms in the lowest energy excited state above the ground state emit these red photons.
What is the difference in energy (J) between the ground state of a Li atom and its lowest-energy
excited state?
d) (1 mark) What is the electron configuration of a ground-state Li atom?
e) (1 mark) What is the electron configuration of a Li atom in its lowest-energy excited state?
f) (1 mark) What region of the
How would you do c and e ? Thanks in advance
c.
delta E = hc/lambda
e.
1s2 2s1 is the ground state.
delta E = 2.180E-18(1/1^2 - 1/x^2)
Solve for x and that is the energy level of the excited state giving the red light. When you know x, then 1s2 2x1. My guess without doing any of the calculation is 2p. The selection rule in spectroscopy is that delta l = +/- 1 so if ground state of the 2s1 electron is 0 (s electrons l = 0) then+1 would be to a p orbital where l = 1.
Post your work if you get stuck.
To answer part (c) of the question, we need to find the energy difference between the ground state of a Li atom and its lowest-energy excited state. The energy difference can be calculated using the equation:
ΔE = E_excited - E_ground
where ΔE is the energy difference, E_excited is the energy of the excited state, and E_ground is the energy of the ground state.
To find the energy of the ground state, we can refer to the electron configuration of a Li atom. The electron configuration of a ground-state Li atom is 1s^2 2s^1, indicating that it has 2 electrons in the 1s orbital and 1 electron in the 2s orbital.
The energy of the lowest-energy excited state can be found by adding one electron to a higher energy level. In this case, to reach the lowest-energy excited state, we add one electron to the 2p orbital. Therefore, the electron configuration of a Li atom in its lowest-energy excited state is 1s^2 2s^1 2p^1.
To calculate the energy difference, we need to know the specific energy values associated with each electron configuration. These values can be obtained from a periodic table or other reference sources.