office hrs - icecube neutrino observatorytmontaruli/phys248/lectures/lecture26.pdf · x-ray spectra...
TRANSCRIPT
Office Hrs
• W 2:10-3:10 pm• F after lecture
• [email protected]• http://www.icecube.wisc.edu/~tmontaruli• Chamberlin Hall - room 4112• Tel. +1-608-890-0901
• This week: more on atom• DC circuits• Magnetic Fields and Induction• EM waves• Cosmology• MTE3 25 April
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What’s your view of atoms?
These 2-3 lectures concern:Bohr atom, X-ray spectra, Frank&Hertz exp. (Ch 4 T&L)Schroedinger equation for H atom, Periodic table and Pauli exclusion principle (Ch 7 T&L)All in Ch 36 of T&M
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Hystory of Atoms• Thompson’s classical model (1897)• Problem: charges cannot be in equilibrium
Thin gold foil
α particlesPlanetary model
Positive charge concentrated in the nucleus (∼ 10-15 m)
Electrons orbit the nucleus (r~10-10 m)
Rutherford’s experiment (1911)
Problem1: emission and absorption at specific frequenciesProblem2: electrons on circular orbits radiate
Emission and Absorption spectra
4http://jersey.uoregon.edu/vlab/elements/Elements.html
Emission spectra: produced by gases where the atom do not experience many collisions. Excited unbound atoms make transitions from excited states to lower levels emitting photonsAbsorption spectra: light crosses gas and atoms absorb at characteristic frequencies. Re-emitted light has different frequencies hence dark linesContinuum spectra: collisions broaden lines and individual lines are no more resolved
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Emitting and absorbing lightZero energy
n=1
n=2
n=3n=4
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E1 = −13.612 eV
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E2 = −13.622 eV
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E3 = −13.632 eV
n=1
n=2
n=3n=4
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E1 = −13.612 eV
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E2 = −13.622 eV
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E3 = −13.632 eV
Photon absorbed hf=E2-E1
Photon emittedhf=E2-E1
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Hydrogen spectra• Lyman Series of emission lines given by
Hydrogen
n=2,3,4,..
Use E=hc/λ
Lyman series
R = 1.096776 x 107 /m
For heavy atoms R∞ = 1.097373 x 107 /m
Rydberg-Ritz
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Bohr’s Model of Hydrogen Atom (1913)
Zero energy
n=1
n=2
n=3n=4
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E1 = −13.612 eV
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E2 = −13.622 eV
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E3 = −13.632 eV
Ener
gy
axis
•Postulate 1: Electron moves in circular orbits where it does not radiate (stationary states)
•Postulate 2: radiation emitted in transitions between stationary states
Orbit radius: rn = n2 a0
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En = −13.6n2 eV
•orbital angular momentum quantized L = mvr = n h/2π
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Ei − E f = hν
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Quantization in physics
“correspondence principle” quantum mechanics must agree with classical results
when appropriate (large orbits and energies)
Incorporating wave nature of electron gives an intuitive understanding of ‘quantized orbits’
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Resonances of a string
Fundamental, wavelength 2L/1=2L, frequency f
1st harmonic, wavelength 2L/2=L, frequency 2f
2nd harmonic, wavelength 2L/3,frequency 3f
λ/2
λ/2
λ/2n=1
n=2
n=3
n=4
freq
uenc
y
. .
.
Vibrational modes equally spaced in frequency
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λn =2Ln
H atom question
Peter Flanary’s sculpture ‘Wave’ outside Chamberlin What quantum state of H?
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Integer number of wavelengths around circumference.
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L = pr = n h2π
⇒hλ
= n h2πr
⇒ 2πr = nλ
Radius and Energy levels of H-atom
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F = k Ze2
r2= m v 2
rmvr = nh
Total energy:
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E =p2
2m−kZe2
r⇒ E = −
12k 2Z 2me4
n2h2= −13.6eV Z 2
n2
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r = n2 h2
mkZe2
=
n2
Za0
This formula agrees to 6 significant digits. For better agreement we have to consider the ‘reduced mass’:
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µ =m
1+ m /M
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EK =p2
2m+
p2
2mN
=12p2 m + M
mM
=
p2
2µ
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pe = pN = p
X-ray spectra
• observed when bombarding target element with high energy electrons in an X-ray tube
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L(n=2)->K(n=1)
M(n=3)->K(n=1)
• When an electron is extracted from the inner shell and an outer electron fills the leftover vacant state, photons are emitted at specific frequencies
• Moseley measured for many elements
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Kα
X-ray spectra
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ν = A(Z − b)
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E2 − E1 = hν ⇒ −13.6eVZ 2
n21n2
−1
Franck and Hertz experiment (1914)
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4.9eV=E1-E0 6.7eV=E2-E0
electrons accelerated up to the energycorresponding to the energy difference between the n level and the fundamental level lose energy in inelastic collisionswith Hg atoms or there can be multipleinelastic collisions
V0