Oh, the humanity! David J. Starling Penn State Hazleton PHYS 214

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1 Oh, the humanity! -Herbert Morrison, radio reporter of the Hindenburg disaster David J. Starling Penn State Hazleton PHYS 24

2 The hydrogen atom is composed of a proton and an electron with potential energy: U(r) = 4πɛ 0 e 2 r.

3 The hydrogen atom is composed of a proton and an electron with potential energy: U(r) = 4πɛ 0 e 2 r. Now, our well is not square but follows an inverse law.

4 All we need to do is solve Schrödinger s equation in spherical coordinates with this radial potential.

5 All we need to do is solve Schrödinger s equation in spherical coordinates with this radial potential. 2 2µ [ r 2 sin θ ( r 2 ψ ) + ( sin θ ψ sin θ r r θ θ sin θ 2 ψ φ 2 ) + ] e2 4πɛ 0 r ψ = Eψ

6 All we need to do is solve Schrödinger s equation in spherical coordinates with this radial potential. 2 2µ [ r 2 sin θ ( r 2 ψ ) + ( sin θ ψ sin θ r r θ θ sin θ 2 ψ φ 2 ) + ] e2 4πɛ 0 r ψ = Eψ We re not there yet, so let s try a different way...

7 Instead, let s think of the electron as orbiting the proton classically.

8 Instead, let s think of the electron as orbiting the proton classically. Using centripetal acceleration, we get F = ma e 2 ) 4πɛ 0 r 2 = m ( v2 r

9 Based on experimental evidence, he know that the energy levels are quantized.

10 Based on experimental evidence, he know that the energy levels are quantized. So let s guess that the angular momentum l is also quantized: l = rmv = n v = n rm for n =, 2, 3,...

11 Combining our quantized angular momentum assumption with the orbital equation, we get r = 4πɛ 0 2 me 2 n 2 = ɛ 0h 2 πme 2 n2

12 Combining our quantized angular momentum assumption with the orbital equation, we get r = 4πɛ 0 2 me 2 n 2 = ɛ 0h 2 πme 2 n2 We call the multiplier of n 2 the Bohr Radius a. and a pm. r = an 2 with a = ɛ 0h 2 πme 2.

13 Combining our quantized angular momentum assumption with the orbital equation, we get r = 4πɛ 0 2 me 2 n 2 = ɛ 0h 2 πme 2 n2 We call the multiplier of n 2 the Bohr Radius a. and a pm. r = an 2 with a = ɛ 0h 2 πme 2. This result is surprisingly accurate!

14 To find the energy, we combine kinetic and potential and then use the orbital equation. e 2 4πɛ 0 r 2 = m v2 r E = 2 mv2 e 2 4πɛ 0 r = e 2 8πɛ 0 r 2

15 To find the energy, we combine kinetic and potential and then use the orbital equation. e 2 4πɛ 0 r 2 = m v2 r E = 2 mv2 e 2 4πɛ 0 r = e 2 8πɛ 0 r 2 Subbing in the Bohr radius formula for r, we get: E n = me4 8ɛ 0 h 2 n 2 = E 0 for n =, 2, 3,... n2 where E 0 = J.

16 To find the energy, we combine kinetic and potential and then use the orbital equation. e 2 4πɛ 0 r 2 = m v2 r E = 2 mv2 e 2 4πɛ 0 r = e 2 8πɛ 0 r 2 Subbing in the Bohr radius formula for r, we get: E n = me4 8ɛ 0 h 2 n 2 = E 0 for n =, 2, 3,... n2 where E 0 = J. This result agrees with quantum theory!

17 The hydrogen atom can absorb a photon if its energy matches an electron transition energy. E γ = hf = E high E low ( ) hc = E 0 λ n 2 high n 2 low ( ) λ = E 0 hc λ = R ( n 2 low n 2 high n 2 high n 2 low with R = m (Rydberg Constant). )

18 The hydrogen atom can absorb a photon if its energy matches an electron transition energy. E γ = hf = E high E low ( ) hc = E 0 λ n 2 high n 2 low ( ) λ = E 0 hc λ = R ( n 2 low n 2 high n 2 high n 2 low with R = m (Rydberg Constant). This is how we find the wavelength of photons emitted from electronic transitions. )

19 Each transition has a unique energy with a photon of a different wavelength.

20 Each transition has a unique energy with a photon of a different wavelength. If the ending state is n =, then that series of wavelengths is known as the Lyman series.

21 But the electron may end up in another state instead.

22 Remember this? 2 2µ [ r 2 sin θ ( r 2 ψ ) + ( sin θ ψ sin θ r r θ θ sin θ 2 ψ φ 2 ) + ] e2 4πɛ 0 r ψ = Eψ

23 Remember this? 2 2µ [ r 2 sin θ ( r 2 ψ ) + ( sin θ ψ sin θ r r θ θ sin θ 2 ψ φ 2 ) + ] e2 4πɛ 0 r ψ = Eψ The ground state solution is: ψ(x) = πa 3/2 e r/a

24 The hydrogen atom s electron, in the ground state, can exist at all radii (except 0). P(r) = 4πr 2 ψ(x) 2 = 4 a 3 r2 e 2r/a

25 The hydrogen atom s electron, in the ground state, can exist at all radii (except 0). P(r) = 4πr 2 ψ(x) 2 = 4 a 3 r2 e 2r/a The peak of this function is at r = a, the Bohr radius!

26 Higher energy states of the hydrogen atom get more complex due to the quantized angular momentum.

27 Higher energy states of the hydrogen atom get more complex due to the quantized angular momentum. These states all have the same energy (n = 2) but different angular momenta (l = 0, ).

28 There are three quantum numbers for the hydrogen atom.

29 There are three quantum numbers for the hydrogen atom.

30 Lecture Question 8. Which of the following most closely resembles the Bohr model of the hydrogen atom? (a) A solid metal sphere with a net positive charge. (b) A hollow metal sphere with a net negative charge. (c) A tray full of mud with pebbles uniformly distributed throughout. (d) The Moon orbiting the Earth. (e) Two balls, one large and one small, connected by a spring.

General Physics (PHY 2140)

General Physics (PHY 2140) General Physics (PHY 140) Lecture 33 Modern Physics Atomic Physics Atomic spectra Bohr s theory of hydrogen http://www.physics.wayne.edu/~apetrov/phy140/ Chapter 8 1 Lightning Review Last lecture: 1. Atomic