Energy transition of LiCl from flame test

AI Thread Summary
The discussion centers on the energy transition of lithium chloride (LiCl) as observed in a flame test, specifically the red color emitted. Participants explore the emission process and the energy level transitions, debating whether the transition occurs from n=2 to n=1 or n=3 to n=2. The Balmer series is referenced, indicating that transitions to n=2 produce visible light, but the applicability of hydrogen's energy levels to lithium is questioned due to lithium's three electrons. There is confusion regarding the lowest energy transition and the relevance of sublevels in this context. Ultimately, the group emphasizes the need to calculate whether the n=2 to n=1 transition falls within the red wavelength range.
Jpyhsics
Messages
84
Reaction score
2

Homework Statement


LiCl flame test is red.
a) define what process
b)and transition between which energy levels?2. The attempt at a solution
a)I believe that the process would be emission, as photons are emitted when energy is released
b)n=2 to n=1
I assumed that that would be the lowest energy transition, which would give us the greatest wavelength of visible light. My only question is why in the hydrogen spectrum is the transition from n=3 to n=2 red light, or anything to transition to n=2 is visible, whereas here would it be different as lithium has 3 e-? Would lithium also emit visible light from transition to n=2? So would n=3 to n=2 be the correct answer?
 
Physics news on Phys.org
Jpyhsics said:
b)n=2 to n=1
I assumed that that would be the lowest energy transition,
Is your "assumption" based on fact; small quantum numbers imply larger or smaller energy changes?
 
Bystander said:
Is your "assumption" based on fact; small quantum numbers imply larger or smaller energy changes?
Well in our lecture our prof said that a small energy change is equal to small frequency and longer wavelength, and I know red is a longer wavelength in the visible spectrum.
 
Jpyhsics said:
Well in our lecture our prof said that a small energy change is equal to small frequency and longer wavelength, and I know red is a longer wavelength in the visible spectrum.
Would n=2 to n=1 be correct?
 
Lyman...Balmer...Paschen...Brackett ...Pfund...? Ring any bells?
 
So far nobody asked: what is Z for lithium?
 
Bystander said:
Lyman...Balmer...Paschen...Brackett ...Pfund...? Ring any bells?
so would it be n=3 to n=2? since the balmer series is visible light and all transitions to n=2?
 
Borek said:
So far nobody asked: what is Z for lithium?
The Z is 3
 
So can you apply hydrogen atom series directly to the energies of the lithium atom transitions?
 
  • #10
Borek said:
So can you apply hydrogen atom series directly to the energies of the lithium atom transitions?
No...?
 
  • #11
Jpyhsics said:
No...?
So what would I have to use?
 
  • #12
Jpyhsics said:
the balmer series
is for hydrogen. Think transition energy.
 
  • #13
Bystander said:
is for hydrogen. Think transition energy.
Well energy wise I am thinking that n=2 to n=1 transition would emit the least amount of energy, which would indicate a longer wavelength. So would that be right?
 
  • #14
Jpyhsics said:
n=2 to n=1 transition would emit the least amount of energy,
"Least?" :oldsurprised:
 
  • #15
"Red" is a reasonably well defined range of wavelengths, calculate if the 2→1 transition falls in that range.
 
  • #16
Borek said:
"Red" is a reasonably well defined range of wavelengths, calculate if the 2→1 transition falls in that range.
Or would this have to do with the sublevels? so the electron goes from 2p to 2s?
 
  • #17
I am a little confused at what everyone is trying to get at.
 
  • #18
Bystander said:
"Least?" :oldsurprised:
Then what would be considered the lowest energy transition?
 
  • #19
Jpyhsics said:
the lowest energy transition?
The "last" Δn=1 transition before the "continuum," ∞-1 to ∞-2, change in energy levels is infinitesimal.:wink:
 
  • #20
I told you - rather precisely - what to check. Have you tried?
 
  • #21
Borek said:
I told you - rather precisely - what to check. Have you tried?
Im not sure the bohr equation would work since it is for atoms with a single electron.
 
Back
Top