Office Hrs

• W 2:10-3:10 pm• F after lecture

• tmontaruli@icecube.wisc.edu• 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