Question 1Level 2 · ExamNumerical answer
A free hydrogen atom, initially at rest, makes the transition n=2→n=1 and emits one photon. Take En=−13.6n2Z2 eV for a hydrogen-like species of nuclear charge Z. Treat the photon energy as equal to the energy difference of the two levels (neglect the tiny recoil correction), and use the photon momentum p=E/c. Take the mass of the hydrogen atom =1.67×10−27 kg, c=3.00×108 m s−1 and 1 eV =1.602×10−19 J. What is the recoil speed of the atom, in m s−1? Give the answer correct to two decimal places.
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Answer: 3.26
Photon energy: E=13.6(1−41)=10.2 eV =10.2×1.602×10−19=1.634×10−18 J. A photon carries momentum p=E/c (the particle side of wave-particle duality). Momentum conservation for the atom initially at rest: Mv=E/c, so v=McE=1.67×10−27×3.00×1081.634×10−18=3.26 m s−1. (The recoil kinetic energy, about 9×10−27 J, is negligible compared with E, which justifies the approximation.)
Difficulty: HardQuestion 2Level 3 · AdvancedNumerical answer
A hydrogen-like ion of unknown nuclear charge Z is initially in an excited level n. It returns to the ground state by emitting exactly two photons, one after the other (that is, via exactly one intermediate level). The two photon energies are 10.2 eV and 108.8 eV, but the order in which they are emitted is not stated. Take En=−13.6n2Z2 eV for a hydrogen-like species of nuclear charge Z. What is the minimum energy (in eV) that would have been needed to ionise the ion from its initial level n? Give the answer correct to two decimal places.
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Answer: 3.4
Let the path be n→m→1. The second photon is E(m→1)=13.6Z2(1−m21). Case 1: the 10.2 eV photon is m→1: Z2(1−1/m2)=0.75 gives Z=1, m=2; then n→2 would have to give 108.8 eV, but in H every transition into n=2 is below 3.4 eV. Impossible. Case 2: 108.8 eV is m→1: Z2(1−1/m2)=8. m=2: Z2=32/3 (no); m=3: Z2=9, Z=3; for m≥4, Z2=8m2/(m2−1) lies strictly between 8 and 9 (no integer). So Z=3 (Li2+), m=3. First photon: 122.4(91−n21)=10.2⇒n21=91−121=361, n=6. Ionisation energy from n=6: 13.6×369=3.40 eV. (Check: 3.40+10.2+108.8=122.4 eV = ground-state ionisation energy of Li2+.) Note that the 10.2 eV photon has the same energy as H Lyman-α, which tempts one to put it last.
Difficulty: Very hardQuestion 3Level 3 · AdvancedOne or more correct
For the hydrogen atom, take λ1=RH(n121−n221) with RH=1.097×107 m−1, where the Lyman, Balmer, Paschen, Brackett and Pfund series have n1=1,2,3,4,5 respectively and n2>n1 (including the series limit n2→∞). Which of the following statements is/are correct? One or more options may be correct.
AThe wavelength ranges of the Lyman and Balmer series do not overlap.
BThe wavelength ranges of the Balmer and Paschen series overlap.
CThe shortest wavelength of the Brackett series is shorter than the longest wavelength of the Paschen series.
DEvery line of the Pfund series has a longer wavelength than every line of the Brackett series.
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Answer: A, C
Range of a series: from the limit λmin=n12/RH to the first line λmax=1/[RH(1/n12−1/(n1+1)2)]. Lyman: 91.2 to 121.5 nm. Balmer: 364.6 to 656.3 nm. Paschen: 820.4 to 1875.2 nm. Brackett: 1458.5 to 4051.5 nm. Pfund: 2278.9 to 7458.4 nm. (A) Lyman ends at 121.5 nm, Balmer starts at 364.6 nm: no overlap. Correct. (B) Balmer ends at 656.3 nm, Paschen starts at 820.4 nm: no overlap. Incorrect. (C) Brackett limit 1458.5 nm < Paschen first line 1875.2 nm. Correct (these two series overlap). (D) Pfund limit 2278.9 nm is shorter than the Brackett first line 4051.5 nm, so Pfund and Brackett overlap; the statement is incorrect. Answer: A, C. (Overlap first appears from Paschen/Brackett onward because the series limits n12/RH grow as n12 while each series spreads over a factor that also grows with n1.)
Difficulty: Very hard