Question 1Level 2 · ExamSingle correct
Use the Bohr model of a one-electron (hydrogen-like) species with an infinitely heavy, point nucleus: En=−13.6Z2/n2 eV. Compare the electron in the n=2 orbit of He+ (Z=2) with the electron in the n=1 orbit of a hydrogen atom (Z=1). Relative to the hydrogen electron, the He+ electron has ______. Exactly one option is correct.
Athe same orbital speed, the same total energy, twice the orbit radius and the same angular momentum
Bthe same total energy, twice the orbit radius and four times the orbital period
Cthe same orbital speed, the same total energy, twice the orbit radius and twice the orbital period
Dthe same orbit radius, the same orbital speed and four times as negative a total energy
Show answer and solution
Answer: C
Bohr-model scalings: rn∝n2/Z, vn∝Z/n, En∝−Z2/n2, Tn=2πrn/vn∝n3/Z2, Ln=nℏ.
He+, n=2 against H, n=1: r: 4/2=2; v: 2/2=1; E: 4/4=1; T: 8/4=2; L: 2ℏ against ℏ, i.e. doubled.
So speed and energy are equal, the radius and period are doubled, and the angular momentum is doubled. Only (C) is fully right: (A) wrongly keeps L unchanged, (B) has T four times, (D) has the radius unchanged.
Difficulty: HardQuestion 2Level 2 · ExamSingle correct
A gas contains identical hydrogen-like ions (atomic number Z not known), all excited to the same level n. Their emission spectrum, made of all possible downward transitions with each distinct wavelength counted once, contains exactly 6 lines, and the line of longest wavelength has photon energy 2.64 eV. Use the Bohr model of a one-electron (hydrogen-like) species with an infinitely heavy, point nucleus: En=−13.6Z2/n2 eV. Take hc=1240 eVnm. The shortest wavelength in this emission spectrum is ______. Exactly one option is correct.
A24.3 nm
B30.4 nm
C22.8 nm
D97.2 nm
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Answer: A
A gas in level n gives (2n) lines (all distinct for one species), so n(n−1)/2=6, n=4.
The longest wavelength is the smallest energy gap, 4→3: 13.6Z2(91−161)=0.661Z2 eV=2.64 eV, so Z2=4, Z=2 (He+).
The shortest wavelength is 4→1: E=54.4×1615=51.0 eV, λ=1240/51.0=24.3 nm.
Traps: 30.4 nm is He+ 2→1; 22.8 nm is the series limit, which is not a line of this spectrum; 97.2 nm is the hydrogen 4→1 line.
Difficulty: HardQuestion 3Level 3 · AdvancedSingle correct
A tungsten target is used in an X-ray (Coolidge) tube. Take the binding energies of K, L and M electrons in tungsten as 69.5 keV, 11.3 keV and 2.3 keV respectively. A vacancy in a shell can be created only if the bombarding electron has at least that shell's binding energy, and a characteristic photon emitted when an electron drops from shell X into a vacancy in shell Y has energy equal to (binding energy of Y) − (binding energy of X). Kα: L→K, Kβ: M→K, Lα: M→L. Take hc=1240 keVpm. The tube is run first at 80 kV and then at 60 kV. Which statement correctly describes the observed spectra? Exactly one option is correct.
AAt 60 kV the Kα line at 21.3 pm appears on the continuum, which begins at 20.7 pm.
BAt 80 kV the continuum begins at 15.5 pm and the Kβ (18.5 pm) and Kα (21.3 pm) lines appear on it; at 60 kV the K lines are absent but the Lα line (138 pm) is still present.
CAt 80 kV the continuum begins at 15.5 pm, the Kα line is at 18.5 pm and the Kβ line is at 21.3 pm.
DAt 60 kV no characteristic line of tungsten appears at all, because 60 kV is below 69.5 kV.
Show answer and solution
Answer: B
Photon energies: Kα =69.5−11.3=58.2 keV (λ=1240/58.2=21.3 pm); Kβ =69.5−2.3=67.2 keV (18.5 pm); Lα =11.3−2.3=9.0 keV (138 pm).
Cut-off: λmin=hc/(eV): 1240/80=15.5 pm at 80 kV, 1240/60=20.7 pm at 60 kV.
At 80 kV the electrons (80 keV>69.5 keV) can create K vacancies, so Kβ and Kα sit on the continuum that starts at 15.5 pm.
At 60 kV, 60<69.5: no K vacancy can be made, so no K line appears, even though 60 keV exceeds the Kα photon energy 58.2 keV. L vacancies (11.3 keV<60 keV) are still created, so the L lines (Lα at 138 pm) remain. That is option (B).
(A) uses the wrong threshold (Kα photon energy instead of the K binding energy); (C) swaps Kα and Kβ (Kβ is the more energetic, shorter line); (D) forgets the L series.
Insight: the threshold for a series is the binding energy of the inner shell, not the energy of the emitted line.
Difficulty: Very hard