(For mobile users) 00:07 Collecting array of microstates into one term (⁴F) 02:23 Collecting array of microstates into second term (²H) 03:21 Collecting array of microstates into third term (²G) 04:27 Collecting array of microstates into fourth term (²F) 05:28 Collecting array of microstates into fifth term (²D) 06:37 Collecting array of microstates into sixth term (⁴P) 07:42 Collecting array of microstates into seventh term (²D, again) 08:47 Collecting array of microstates into eighth term (²P) 09:44 List of terms for d³ 10:55 Using Hund's (three) Rules to find ground state 11:39 The quantum number J
⁴𝘍₃/₂ is the ground state; all of the other terms - ⁴P, ²𝘏, ²𝘎, ²𝘍, ²𝘋, ²𝘗, and the other J-values for ⁴𝘍 (⁴𝘍 ₃/₂, ₅/₂, ₇/₂) are excited states. Note that, while we use Hund's three (3) rules to find the ground state (here, ⁴𝘍₃/₂), we cannot use them to put the excited states in order of energy. (For example, we cannot tell which excited state would the 1st excited, 2nd excited , and so on). Does that make sense?
Not for d2, but here is p2, if that might be helpful: ruclips.net/video/mXc4E4Zg5YE/видео.html ruclips.net/video/twasJ1XeY_s/видео.html ruclips.net/video/BcJLcU_Z1Tw/видео.html
Great question! To have a ⁴D term, we would need at least one microstate for each combination of MS = (-3/2, -1/2, +1/2, and +3/2), AND ML = (-2, -1, 0, +1, and +2). If we don't have all 20 of those specific microstates (4 MS values x 5 ML values), we don't have a ⁴D term. One way to visualize this is to notice that the "tick marks" (little lines) make a (filled) rectangle for each spectroscopic term. I have shown this "rectangle" using different colored lines. Does that help?
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(For mobile users)
00:07 Collecting array of microstates into one term (⁴F)
02:23 Collecting array of microstates into second term (²H)
03:21 Collecting array of microstates into third term (²G)
04:27 Collecting array of microstates into fourth term (²F)
05:28 Collecting array of microstates into fifth term (²D)
06:37 Collecting array of microstates into sixth term (⁴P)
07:42 Collecting array of microstates into seventh term (²D, again)
08:47 Collecting array of microstates into eighth term (²P)
09:44 List of terms for d³
10:55 Using Hund's (three) Rules to find ground state
11:39 The quantum number J
you saved my academic life. Thanks!
Happy to help!
Thank you so much sir. You have clarified my doubt
Thanks
Thank you so much!
Glad it helped!
Sir from this we can say that for "d3" 4F will we the ground energy term, then what will be it's excited energy term?
⁴𝘍₃/₂ is the ground state; all of the other terms - ⁴P, ²𝘏, ²𝘎, ²𝘍, ²𝘋, ²𝘗, and the other J-values for ⁴𝘍 (⁴𝘍 ₃/₂, ₅/₂, ₇/₂) are excited states.
Note that, while we use Hund's three (3) rules to find the ground state (here, ⁴𝘍₃/₂), we cannot use them to put the excited states in order of energy. (For example, we cannot tell which excited state would the 1st excited, 2nd excited , and so on).
Does that make sense?
How to write F,
Sir can u give any link for d2
Not for d2, but here is p2, if that might be helpful:
ruclips.net/video/mXc4E4Zg5YE/видео.html
ruclips.net/video/twasJ1XeY_s/видео.html
ruclips.net/video/BcJLcU_Z1Tw/видео.html
Why isn't 4D a possible term? MS = 3/2 . ML = (+2, +1, -1)
Great question!
To have a ⁴D term, we would need at least one microstate for each combination of MS = (-3/2, -1/2, +1/2, and +3/2), AND ML = (-2, -1, 0, +1, and +2). If we don't have all 20 of those specific microstates (4 MS values x 5 ML values), we don't have a ⁴D term.
One way to visualize this is to notice that the "tick marks" (little lines) make a (filled) rectangle for each spectroscopic term. I have shown this "rectangle" using different colored lines.
Does that help?
@@lseinjr1 Yes thank you so much, I thought if only one combination was possible then the entire term was possible, this helps so much :)