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Questions on energy levels and photon emission
A glowing gas emits a line spectrum rather than a continuous one, evidence that the electrons in an atom occupy discrete energy levels. One electron falling between two levels emits one photon carrying exactly the gap, hf = E₁ − E₂, and each element's own set of levels gives it a characteristic line spectrum.
19 original questions · 57 marks · the energy levels and photon emission notes · Quantum phenomena
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Explain what the sharp lines of an emission line spectrum tell us about the energy of electrons in an atom.
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Each line corresponds to a photon of a definite energy, emitted when an electron falls between two levels (1). Sharp, discrete lines show the electron energies are quantised (only certain values are allowed), not continuous (1).An electron transition releases 3.0 eV of energy. Calculate this in joules (e = 1.60 × 10−19 C).
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3.0 eV = 3.0 × 1.60 × 10−19 (1)
= 4.80 × 10−19 J (1)State the equation relating the frequency of an emitted photon to the energy levels involved.
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hf = E1 − E2 (1), where E1 and E2 are the higher and lower energy levels of the transition (1).State what is meant by the ground state of an atom.
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The lowest energy level of the atom (1).The energy levels of an atom are given negative values. Explain why.
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The zero of energy is taken as an electron just free of the atom, at ionisation (1). A bound electron has less energy than a free one, since energy must be supplied to remove it, so every level lies below zero (1).White light is passed through a cool gas and then through a diffraction grating. Dark lines appear in the otherwise continuous spectrum. Explain why.
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The gas absorbs only photons whose energy exactly matches a gap between two of its energy levels (1). Light of those frequencies is removed from the beam (and re-emitted in all directions), leaving dark lines at those positions (1).An electron falls from a level at −1.5 eV to a level at −3.4 eV. Taking h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1 and e = 1.60 × 10−19 C, calculate (a) the photon frequency and (b) its wavelength.
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ΔE = (−1.5) − (−3.4) = 1.9 eV (1)
ΔE = 3.04 × 10−19 J (1)
(a) f = ΔE/h = 4.59 × 1014 Hz (1)
(b) λ = c/f = 654 nm (1)A transition releases a photon of energy 2.55 eV. Calculate its wavelength.
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λ = hc/ΔE (1)
= (6.63 × 10−34 × 3.0 × 108)/(2.55 × 1.60 × 10−19) (1)
λ = 488 nm (1)An electron drops from a level at −0.54 eV to a level at −3.40 eV. Calculate the wavelength of the emitted photon.
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ΔE = (−0.54) − (−3.40) = 2.86 eV (1)
λ = hc/ΔE (1)
λ = 435 nm (1)A photon of wavelength 486 nm is emitted from an atom. Calculate the energy of the transition, in electronvolts.
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E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(486 × 10−9) (1)
E = 4.09 × 10−19 J (1)
E = 2.56 eV (1)The red line in the spectrum of atomic hydrogen has a wavelength of 656 nm and is produced by electrons falling to the level at −3.40 eV. Determine the energy of the level from which the electrons fall (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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Photon energy E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(656 × 10−9) = 3.03 × 10−19 J (1) = 1.90 eV (1). Upper level = −3.40 + 1.90 = −1.50 eV (1).A sodium atom has its ground state at −5.14 eV and its first excited level at −3.04 eV. Calculate the frequency of the photon that must be absorbed to raise an electron from the ground state to the first excited level (h = 6.63 × 10−34 J s, e = 1.60 × 10−19 C).
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ΔE = (−3.04) − (−5.14) = 2.10 eV = 3.36 × 10−19 J (1). f = ΔE/h (1) = 5.07 × 1014 Hz (1).Four of the energy levels of atomic hydrogen are −0.85 eV, −1.51 eV, −3.40 eV and −13.6 eV (the ground state). A line in the ultraviolet spectrum of hydrogen has a wavelength of 122 nm. Deduce which transition produces this line (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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Photon energy E = hc/λ = (6.63 × 10−34 × 3.0 × 108)/(122 × 10−9) = 1.63 × 10−18 J (1) = 10.2 eV (1). The only gap of this size is (−3.40) − (−13.6) = 10.2 eV (1), so the line comes from electrons falling from the −3.40 eV level to the ground state (1).An atom has levels at 0 eV (ionised), −1.5 eV and −13.6 eV. An electron falls from the −1.5 eV level to the −13.6 eV level. Calculate the wavelength of the photon emitted and state which part of the spectrum it lies in.
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ΔE = (−1.5) − (−13.6) = 12.1 eV (1)
ΔE = 1.94 × 10−18 J (1)
λ = hc/ΔE = 103 nm (1)
which is in the ultraviolet (1)An atom has three energy levels: −1.5 eV, −3.4 eV and −13.6 eV. State how many different spectral lines can be produced by transitions between them, and calculate the wavelength of the line with the longest wavelength.
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Three transitions are possible (3→2, 3→1, 2→1) (1)
the three energy gaps are 1.9 eV, 10.2 eV and 12.1 eV, all different, so the three transitions give three distinct lines (1)
The longest wavelength comes from the smallest energy gap, −1.5 to −3.4 eV (1.9 eV) (1)
λ = hc/ΔE = 654 nm (1)Explain why the line spectrum of each element is unique.
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Every element has its own distinct set of electron energy levels (1), so the differences between levels, and therefore the photon energies and wavelengths emitted (1), form a pattern unique to that element (1).An atom has three energy levels: −0.9 eV, −2.8 eV and −4.6 eV (the ground state). A student claims that all three possible spectral lines lie in the visible range, 400 nm to 700 nm. Deduce whether the claim is correct (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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The three gaps are 1.9 eV, 1.8 eV and 3.7 eV (1). For each, λ = hc/ΔE (1). 1.9 eV gives 654 nm and 1.8 eV gives 691 nm, both within 400 to 700 nm (1). 3.7 eV gives 336 nm, which is ultraviolet (1). The claim is incorrect: two lines are visible but the third is not (1).An atom has energy levels at −1.0 eV, −2.5 eV and −6.0 eV (the ground state). An electron in the −1.0 eV level returns to the ground state via the −2.5 eV level. State the number of photons emitted and calculate the wavelength of each (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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Two photons, one for each downward transition (1). The gaps are 1.5 eV and 3.5 eV (1). λ1 = hc/ΔE = 829 nm (infrared) (1); λ2 = 355 nm (ultraviolet) (1).A hydrogen atom is in its first excited state, at −3.40 eV. Ionisation corresponds to 0 eV. Determine the maximum wavelength of electromagnetic radiation that can ionise the atom from this state (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1, e = 1.60 × 10−19 C).
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The photon must supply at least 3.40 eV = 5.44 × 10−19 J (1). The maximum wavelength goes with this minimum energy: λ = hc/E (1) = 366 nm (1).
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