Physics 228

Today:

Bohr Model of H Atom Schrodinger Equation

Bohr Atom Radius

• Consider an electron in circular orbit about a proton of radius rn. The circumference is 2πrn.

• The electrostatic attraction force between the electron and proton is F = e2/4πε0rn

2.

• In a circular orbit, this force must be equal to the mass times the centripetal acceleration m vn

2/rn.

• Multiply both sides of the force equation by m rn3 to obtain:

m e2 rn/4π ε0 = m2 vn2 rn

2 = Ln2.

• Now we use the quantization of angular momentum: Ln = mevnrn = nħ.

• Solve for rn : rn = n2 h2ε0/πme2 ≡ n2 a0.

• The "Bohr radius" is a0 = 0.053 nm = 53 pm.

Bohr Atom Energy Levels

• Solving for the potential energy in the n’th orbit gives:

Un = -e2/4πε0rn = -mee4/4ε0

2n2h2.

• The kinetic energy is

Kn = p2/2m = (n ħ/rn)2/2m = mee4/8ε0

2n2h2 .

• We see that Kn = -½ Un so the total energy of the hydrogen atom is

En = Kn + Un = -mee4/8ε02n2h2.

• Plugging in numbers for the constants, we obtain

En = -13.6 eV / n2.

Bohr Model Transition Energies

Rydberg Formula for Hydrogen

With En = -13.6 eV / n2, the Bohr model of the hydrogen atom predicts that a transition from an (upper) level “m” to (lower) level “n” will result in emission of a photon of energy

𝐸 = 13.6 𝑒𝑉1

𝑛2−

1

𝑚2.

Despite the simplicity of Bohr’s model, this is exactly (well, almost…) what is observed:

Light emitted by hydrogen gas discharge: Balmer Series

Bohr Model Takeaway

The Bohr atom is based on E&M plus the assumption that circular orbits have an integral number of wavelengths. It leads to:

Quantized radii

Quantized angular momenta

Quantized energies

The energy levels are predicted correctly, and the concept of angular momentum quantization is also correct.

However, the model predicts that the n’th energy level always has an angular momentum of nħ, which is not confirmed experimentally.

Thus, the Bohr model allows us to start understanding quantization, but it is not really right. Full understanding requires the theory of quantum mechanics.

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For a proton to have the same wavelength as an electron

a) it has to have the same momentum as the electron

b) it has to have the same kinetic energy as the electron

c) it has to have the same mass as the electron

d) it has to have the same speed as the electron

e) A proton can never have the same wavelength as an electron.

Wave Equation for Light in 1 Dimension

From Maxwell’s equations, we can derive a wave equation for the electric field:

A light wave can be represented as E = E0 cos(kx - ωt).

Inserting into wave equation we obtain: -k2 E = (1/c2) (-ω2) E

k2 = ω2/c2

c = ω/k.

What is the Wave Equation for Particles (Matter Waves)?

The wave equation for particles has to be different, since for a non-relativistic particle we have kinetic energy E = p2/2m (not E = pc as for light).

We require the de Broglie relations to apply: E = ħω and p = ħk.

Then p2/2m = E ➮ ħ2k2/2m = ħω.

The factor of k2 should look familiar - recall the 2nd derivative with respect to x (previous slide, wave equation left-hand side) for the wave equation for light gave: -k2 E(x,t).

The factor of ω is also familiar - that comes from the 1st derivative of the right hand side of the wave equation.

Wave Equation for Particles

For the particle amplitude let’s use the symbol Ψ. After plugging Ψ into the wave equation, want to get ħ2k2/2m = ħω.

So we should have an equation that relates the second x-derivative to the first time derivative of Ψ:

The -ħ2/2m was put on the left to get the kinetic energy, while the C was put on the right since we are not certain what we will need there.

Let's assume Ψ(x,t) = A cos(kx-ωt) as before.

Then the L.H.S. gives (ħ2k2/2m) Ψ(x,t), as desired.

The R.H.S. gives -ωCA sin(kx-ωt). This has the ω we need, but it cannot work with cos(kx-ωt) on one side and sin(kx-ωt) on the other.

☹ We need something better...

Complex Numbers

• A complex number has the form z = x + iy, where x is the real part, and y is the imaginary part.

• A complex number is represented by a point in the complex plane.

• By expanding exp(z), sin(z), and cos(z) functions into Taylor series, we can show that

real axis

imaginary axis

iy

x

θ

𝑧 = 𝑥 + 𝑖𝑦

𝑧 = 𝑥2 + 𝑦2

𝑧 = 𝑧 cos 𝜃 + 𝑖 sin 𝜃

= 𝑧 𝑒𝑖𝜃

What is the modulus of 𝑒𝑖𝜑 ?

a.) 𝑒𝑖𝜑 = 1

b.) 𝑒𝑖𝜑 depends on the value of 𝜑

c.) 𝑒𝑖𝜑 = 2

d.) 𝑒𝑖𝜑 = 0 e.) The modulus is not defined for a complex argument.

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