A-level Physics (Advancing Physics)/Energy Levels

As an electron approaches a nucleus from infinity, it becomes ‘bound’ – it is attached to the nucleus, if you like. In this bound state, the electron occupies what is called an energy level. A nucleus has a discrete number of energy levels, and so electrons bound to a certain nucleus can only take on certain potential energies. These energies are negative by convention. The lowest (most negative) energy level is denoted n = 1, the next lowest n = 2, and so on. The values of these can be found using formulæ. Alternatively, they may be determined experimentally. thumb|right|200px|Energy levels in a hydrogen atom. The transition shown from the n = 3 level to the n = 2 level gives rise to visible light of wavelength 656 nm nanometres (red). At random, electrons jump between energy levels. If they jump to a lower energy level (more negative), they release energy in the form of a photon. If they jump to a higher energy level, they must absorb a photon of the appropriate energy. The energies of these photons can be calculated using the following formulæ, which you should already know from AS Applied Science: E = h ⋅ f E(γ) c = λ ⋅ f c where E is energy, h is Planck’s constant (6.63 ⋅ 10−34 J joule ⋅ s seconds), f is frequency, c is the speed of light, and λ is wavelength. The energy levels of different nuclei are different. Evidence for these energy levels comes from the emission and absorption spectra of atoms. An emission spectrum can be obtained by heating a sample of an element. This gives the electrons energy, so they jump up the energy levels. At random, they then jump down again, giving off photons with measurable frequencies. The formulæ above can be used to calculate the difference in energy between the levels between which the electrons have jumped. An absorption spectrum can be found by passing light through (for example) a gas, and observing the frequencies of light which are absorbed. These frequencies correspond to jumps between energy levels which electrons have undergone when they absorb the photons, gaining energy. It should be noted that electrons do not always jump to the next‐door energy level – they can, in principle, jump to any energy level. They cannot jump to an energy which is not that of an energy level. To work out how much (electrostatic potential) energy an electron will have at a certain energy level, use the formula: E = −13.6 n2 En To work out how much energy an electron will gain when jumping between energy levels, use the formula: ΔE = −13.6 (n2 1 − n2 2) ΔEn where an electron is transitioning between n1 and n2. The value of 13.6 is a constant by itself but is made up of a combination of constants from the derivation of the Bohr radius and energy levels in hydrogen. 13.6 comes from e3 ⋅ me 8 ⋅ h2 ⋅ ε0 13.6 Try it yourself: charge on an electron, e = 160 zC zeptocoloumb, mass of an electron, me = 911 qg quectogramme, Planck’s Constant, h = 663 aJ attojoule ⋅ as attosecodne, and permittivity of free space, ε0 = 8.85 pC picocoloumb V volt ⋅ m metre. Questions { For automatic marking round your answers to three digits. The following table gives the wavelengths of light given off when electrons change between the energy levels in hydrogen as described in the first row: Transition of n 3 → 2 4 → 2 5 → 2 6 → 2 7 → 2 8 → 2 9 → 2 ∞ → 2 Wavelength (nm nanometres) 656.3 486.1 434.1 410.2 397.0 388.9 383.5 364.6 Colour Red Blue‐green Violet Violet (Ultraviolet) (Ultraviolet) (Ultraviolet) (Ultraviolet) } { Calculate the potential energy of an electron at level n = 2. | type="{}" } { −546 zJ | −546zJ | -546zJ | −5.46 · 10⁻¹⁹ J | −5.46·10⁻¹⁹J | −5.46 × 10⁻¹⁹ J | −5.46×10⁻¹⁹J | −3.41 eV | −3.41eV } || { Calculate the difference in potential energy between levels n = 2 and n = 3. | type="{}" } { 303 zJ | 303zJ | +303 zJ | +303zJ | 3.03 · 10⁻¹⁹ J | 3.03·10⁻¹⁹J | 3.03 × 10⁻¹⁹ J | 3.03×10⁻¹⁹J | 1.89 eV | 1.89eV } || This time, let’s derive a general formula: { What is the potential energy of an electron at level n = 3? | type="{}" } { −1.52 eV | −1.52eV | -1.52eV } || { If an electron were to jump from n = 7 to n = 5, what would the wavelength of the photon given off be? | type="{}" } { 4.65 μm | 4.65μm | 4.65 micrometre | 4.65 micrometres | 4.65 micrometre | 4.65 micrometres | 4.65 · 10⁻⁶ m | 4.65·10⁻⁶m | 0.00000465m } || { Prove that the wavelength of light emitted from the transition n = 4 to n = 2 is 486.1 nm nanometre. (Hint: E(γ) and c) | type="()" } + (self‐marked answer) ||