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10.2.7
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Anyone good at physics? REALLLY need help...=o
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Post by
149424
This post was from a user who has deleted their account.
Post by
kattib
Hmm only one I may know at all (havnt done quantum in physics yet) is a. isnt energy released equal to the quanta dropped times plancks constant times frequency of the electron?
I dunno 100% so sorry
Post by
149424
This post was from a user who has deleted their account.
Post by
Heckler
It's been awhile since Physics, but here's a shot:
====================================
En = -E1 / n^2
E1 == 1 / (4 pi eps0) * e^2 / (2 * aB) where eps0 = permittivity constant, e = fundamental charge, aB = bohr radius; E1 == 13.60 eV
a.) E6 - E3 = 1.817 (10^-19) J =
1.13419 eV
b.) deltaE / h = Frequency of emitted photon. Wavelength = c / freq. So, Wavelength = c*h / deltaE = 1.093 (10^-6) m (1093 nm)
As a check, cookie cutter formula = Lamba0 / (1/m^2 - 1/n^2) where Lambda0 = 91.12nm, n=6, m=3..... =
1093 nm
c.) I'm not 100% on this one, attempt to use above eq: 958nm = 91.12nm / (1/3^2 - 1/x^2) solve for x ==
7.90661
So... 7 or 8? Probably 7, but again, not positive.
d.) Unsure here too, I can't recall how l = 2 changes the rule about L being an integer multiple of h*... but a guess: L = sqrt(6) * h* = 2.57196*10^-34 kg*m/s =
1.60529*10^-15 eV * s
I'll go ahead and post this now in case its helpful, but I'm going to dig up my chemistry text and see if I can verify / correct anything...
First edit:
For part d, I think I gave you the
Orbital
component of Angular Momentum, not the Total Angular momentum... I don't think l=2 affects total angular momentum, I think only n affects that... d might be just n*h* = 1.97463*10^-15 eV*s --- still not sure on this, it seems strange to provide l = 2 if its not to be used (unless he wants the Orbital / Azimuthal vector form or something? That seems rather complex)
Second Edit:
Although I'm still not positive, I think my answer above for D is what your instructor is looking for. sqrt(6) * h*. = 1.60529 * 10^-15 eV*s (which would be the correct answer regardless of the value of n for any d-orbital. which is why I'm still a little unsure.)
Post by
Kibbles
sorry I lost you on "Consider"
Post by
Orranis
It's been awhile since Physics, but here's a shot:
====================================
En = -E1 / n^2
E1 == 1 / (4 pi eps0) * e^2 / (2 * aB) where eps0 = permittivity constant, e = fundamental charge, aB = bohr radius; E1 == 13.60 eV
a.) E6 - E3 = 1.817 (10^-19) J =
1.13419 eV
b.) deltaE / h = Frequency of emitted photon. Wavelength = c / freq. So, Wavelength = c*h / deltaE = 1.093 (10^-6) m (1093 nm)
As a check, cookie cutter formula = Lamba0 / (1/m^2 - 1/n^2) where Lambda0 = 91.12nm, n=6, m=3..... =
1093 nm
c.) I'm not 100% on this one, attempt to use above eq: 958nm = 91.12nm / (1/3^2 - 1/x^2) solve for x ==
7.90661
So... 7 or 8? Probably 7, but again, not positive.
d.) Unsure here too, I can't recall how l = 2 changes the rule about L being an integer multiple of h*... but a guess: L = sqrt(6) * h* = 2.57196*10^-34 kg*m/s =
1.60529*10^-15 eV * s
I'll go ahead and post this now in case its helpful, but I'm going to dig up my chemistry text and see if I can verify / correct anything...
Until now, I was really enthusiastic about taking physics in highschool.
Then again, in 5th grade, I was scared sh*tless by algebra, turned out I actually knew how to do it, without knowing what it actually was, so it might be easier than it looks...
Post by
Monday
It's been awhile since Physics, but here's a shot:
====================================
En = -E1 / n^2
E1 == 1 / (4 pi eps0) * e^2 / (2 * aB) where eps0 = permittivity constant, e = fundamental charge, aB = bohr radius; E1 == 13.60 eV
a.) E6 - E3 = 1.817 (10^-19) J =
1.13419 eV
b.) deltaE / h = Frequency of emitted photon. Wavelength = c / freq. So, Wavelength = c*h / deltaE = 1.093 (10^-6) m (1093 nm)
As a check, cookie cutter formula = Lamba0 / (1/m^2 - 1/n^2) where Lambda0 = 91.12nm, n=6, m=3..... =
1093 nm
c.) I'm not 100% on this one, attempt to use above eq: 958nm = 91.12nm / (1/3^2 - 1/x^2) solve for x ==
7.90661
So... 7 or 8? Probably 7, but again, not positive.
d.) Unsure here too, I can't recall how l = 2 changes the rule about L being an integer multiple of h*... but a guess: L = sqrt(6) * h* = 2.57196*10^-34 kg*m/s =
1.60529*10^-15 eV * s
I'll go ahead and post this now in case its helpful, but I'm going to dig up my chemistry text and see if I can verify / correct anything...
Until now, I was really enthusiastic about taking physics in highschool.
Then again, in 5th grade, I was scared sh*tless by algebra, turned out I actually knew how to do it, without knowing what it actually was, so it might be easier than it looks...
Pretty much this.
Post by
Squishalot
It's easier than it looks, it's just math.
I had an actuarial crib sheet (page of notes I can bring into exams with hints and tips and formulae etc) for an exam, and I was shocked afterwards to realise that my sheet (which was essentially for a maths and statistics exam) had no numbers on it, other than for the power (eg, x-squared, x-cubed etc) numbers. Sucks, but you get used to it after a while.
Post by
Monday
It's easier than it looks, it's just math.
What's your point?
To me those two phrases are mutually exclusive.
Post by
149424
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Post by
150529
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Post by
128491
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