molecular energy storage and particle in a box

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15 Terms

1
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em spectrum (no x-ray or gamma) from lower to higher energy

assign: molecular vibrations, nuclear spins, electronic transitions, molecular rotations and electron spins to which region

label: increasing E, increasing frequency, increasing wavenumber and increasing wavelength

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2
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how are different forms of energy storage treated and why (give an equation)

  • various forms of energy in a molecule can be treated differently

  • this is because they occur on different time scales

  • these energies are quantised and most (except translational) can be measured by spectroscopy

<ul><li><p>various forms of energy in a molecule can be treated differently</p></li><li><p>this is because they occur on different time scales</p></li><li><p>these energies are quantised and most (except translational) can be measured by spectroscopy</p></li></ul><p></p>
3
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x and V for a particle confined to a 1D box of length l

diagram - “y” axis, “x” axis, movement of particle

inside: 0 ≤ x ≤ l and V(x) = 0

outside: x ≤ 0 or x ≥ l. V(x) = ∞ → the particle cannot exist in these regions of space where ψ (x) = 0

<p>inside: 0 ≤ x ≤ l and V(x) = 0</p><p>outside: x ≤ 0 or x ≥ l. V(x) = ∞ → the particle cannot exist in these regions of space where ψ (x) = 0</p>
4
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from previous set: what is the solution (for wavefunction) to the schrodinger equation in regions of constant potential?

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5
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boundary conditions for particle in a box and why these are defined

what does this mean for allowed energies

as ψ (x) = 0 outside the box, and for continuity of the wavefunction, the boundary conditions are defined:

ψ (x) = 0, x = 0, l at the edges

therefore the allowed energies are integer numbers of half wavelengths of the sine wave

6
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1st part of particle in a box derivation using the boundary conditions

at x = 0, ψ = 0. A = 0 is not an allowed condition so sin(kx+ϕ) = 0

x = 0, so Asin(ϕ) = 0

this is satisfied by ϕ = nπ

at x = l, ψ = 0.

Asin(kl) = 0 → sin(kl) = 0 → kl = nπ → k = nπ/l

ψ (x) = Asin((nπ/l)x) when 0 ≤ x ≤ l

<p>at x = 0, ψ = 0. A = 0 is not an allowed condition so sin(kx+ϕ) = 0</p><p>x = 0, so Asin(ϕ) = 0</p><p>this is satisfied by ϕ = nπ</p><p></p><p>at x = l, ψ = 0.</p><p>Asin(kl) = 0 → sin(kl) = 0 → kl = nπ → k = nπ/l</p><p></p><p>ψ (x) = Asin((nπ/l)x) when 0 ≤ x ≤ l</p>
7
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equation for En and how is it derived from particle in a box

<p></p>
8
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<p>how is A found and how can the equation for ψ be expressed as a result?</p>

how is A found and how can the equation for ψ be expressed as a result?

  1. know particle is in box so set probability to 1.

  2. substitute in the equation for ψ

  3. use the identity for sin2(ax)

  4. solve to find A

  5. put A back into equation for ψ

<ol><li><p>know particle is in box so set probability to 1.</p></li><li><p>substitute in the equation for ψ</p></li><li><p>use the identity for sin<sup>2</sup>(ax)</p></li><li><p>solve to find A</p></li><li><p>put A back into equation for ψ</p></li></ol><p></p>
9
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<p>full derivation (only do after know other flashcards)</p>

full derivation (only do after know other flashcards)

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10
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n = 1 to n = 4 particle in a box diagram including equations for E

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11
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how can the equation for En be derived from the de Broglie wavelength

  1. inside the box, the potential energy, V, is zero, so energy is all kinetic

  2. substitute v from kinetic energy into de Broglie equation

  3. use the allowed wavefunctions formulae and sub in wavelength

  4. rearrange to get equation for En - agrees with Schrodinger approach

<ol><li><p>inside the box, the potential energy, V, is zero, so energy is all kinetic</p></li><li><p>substitute v from kinetic energy into de Broglie equation</p></li><li><p>use the allowed wavefunctions formulae and sub in wavelength</p></li><li><p>rearrange to get equation for E<sub>n</sub> - agrees with Schrodinger approach</p></li></ol><p></p>
12
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Etranslation for 3D particle in a box

3D box with sides lx, ly and lz

<p>3D box with sides l<sub>x</sub>, l<sub>y</sub> and l<sub>z</sub></p>
13
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lowest energy transition and approximate ΔE for He atom in a cubic box of side 1m

comment on value of ΔE and what this means

lowest energy transition is from E1,1,1 → E2,1,1 (= E1,2,1 = E1,1,2)

ΔE ≈ 2.48 × 10-41 J/molecule - so tiny and so transitions between translational energy levels are too small to measure, meaning translational motion can be treated as a classical continuum

14
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what are quantum dots

semiconducting, fluorescent nanoparticles ranging from ~2-10 nm in diameter. they emit visible light. electrons are confined to the particle so energy levels can be modelled using 3D particle in a box

15
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<p>emission wavelength, <strong>Δ</strong>E and particle size from left to right</p>

emission wavelength, ΔE and particle size from left to right

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