Bohr orbit radius: $r_n = n^2 \times 0.529 \times 10^{-10}$ m
For $n=1$: $r_1 = 0.529\times10^{-10}$ m
In units of $10^{-10}$ m (Angstrom): $r_1 = \mathbf{0.529}$ Angstrom
Answer: 0.529
Bohr (1913) postulates: electrons move in circular orbits, angular momentum $= nh/2\pi$ (quantised), no radiation while in orbit, radiation on transition. Results: $r_n = n^2 a_0$ ($a_0=0.529$ A), $v_n = e^2/4\pi\varepsilon_0\hbar n$, $E_n = -13.6/n^2$ eV. Emission: photon emitted when electron falls to lower orbit. Absorption: photon absorbed when electron jumps to higher orbit.
Wavelength from transition $n_i \to n_f$: $1/\lambda = R_H(1/n_f^2 - 1/n_i^2)$ where $R_H = 1.097\times10^7$ m$^{-1}$. Lyman: $n_f=1$, UV range. Balmer: $n_f=2$, visible (H$\alpha=656$ nm, H$\beta=486$ nm, H$\gamma=434$ nm, H$\delta=410$ nm). Paschen: $n_f=3$, near IR. Convergence limit: as $n_i\to\infty$, lines crowd together. Series limit wavelength: $\lambda_{limit} = n_f^2/R_H$.
Each element has unique emission/absorption spectrum — atomic fingerprint. Emission spectrum: heated gas emits specific wavelengths → bright lines on dark background. Absorption spectrum: white light through cool gas → dark lines at same wavelengths → Fraunhofer lines in solar spectrum. Spectroscopy used: stellar composition analysis (Fraunhofer lines), identifying elements in flames (flame test), forensic analysis, medical diagnostics (NMR/MRI based on nuclear spin resonance).
Characteristic X-rays: inner shell electrons knocked out by high-energy electrons in X-ray tube → outer electrons fall in → emit X-rays at specific energies. Bremsstrahlung: continuous X-ray spectrum from electron deceleration. Moseley\'s law: $\sqrt{f} \propto (Z - b)$ where $Z$ = atomic number — explained by Bohr model (confirmed atomic number ordering). X-ray diffraction (Bragg\'s law): $2d\sin\theta = n\lambda$; used for crystal structure determination. Compton scattering: X-ray + electron → longer wavelength X-ray, proved photon momentum.
The radius of the nth orbit is $r_n = \frac{n^2}{Z} \times 0.529$ Å, so it grows as $n^2$ and shrinks as the nuclear charge rises. The velocity is $v_n = \frac{Z}{n} \times 2.18 \times 10^6$ m/s, falling as 1/n. Combining them gives the time period, which scales as $n^3/Z^2$. Most numerical questions in this topic reduce to identifying which of these three scalings the question is testing.
The general result is $E_n = -13.6 \frac{Z^2}{n^2}$ eV, so He⁺ (Z = 2) has levels four times deeper than hydrogen and Li²⁺ (Z = 3) nine times deeper. The ionisation energy is the energy needed to reach $n = \infty$, which is simply $13.6Z^2$ eV from the ground state. The negative sign means the electron is bound; a positive total energy would mean it has escaped.
The model works only for one-electron systems — it fails for helium onward because it ignores electron-electron repulsion. It also cannot explain the relative intensities of spectral lines, the fine structure revealed by high-resolution spectroscopy, or the Zeeman splitting of lines in a magnetic field. Most fundamentally, it assumes a definite orbit, which the uncertainty principle forbids. Quantum mechanics replaced the orbit with an orbital — a probability distribution.
Forgetting the $Z^2$ factor for ions. Using $-13.6/n^2$ for He⁺ gives an answer four times too small.
Mixing up the series. A series is named by the level the electron lands on, not the one it starts from. Only the Balmer series falls in the visible range.