Bohr Model Numericals
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Direct answer
Bohr's 1913 model quantises the hydrogen atom into orbits of angular momentum mvr = nh/2π, and from it flow the three results NCERT expects verbatim: radius r_n = 0.529 n² Å, energy E_n = −13.6/n² eV, and for any transition 1/λ = R(1/n1² − 1/n2²) with R = 1.097 × 10^7 m^-1. The negative energy marks a bound system; the ground state sits at −13.6 eV, so ionisation from the ground state costs exactly 13.6 eV, and from n = 2 only 3.4 eV. Every NEET numerical in this block is a transition wavelength, an energy difference, or an orbit property scaled by n² — three formulas dressed in endless costumes.
What you must remember
- Quantisation postulate: L = nh/2π, n = 1, 2, 3...; from it, r_n = 0.529 n² Å and v_n = 2.18 × 10^6/n m/s (about c/137 in the first orbit).
- Energy ladder: E_n = −13.6/n² eV; the rungs NEET recycles are E1 = −13.6, E2 = −3.4, E3 = −1.51, E4 = −0.85 eV.
- Rydberg formula: 1/λ = R(1/n1² − 1/n2²), R = 1.097 × 10^7 m^-1; Hα (3 → 2) at 656 nm, Balmer limit 364.6 nm, Lyman limit 91.2 nm.
- Series placement: Lyman ends on n = 1 (ultraviolet); Balmer on n = 2 (visible — the only visible series); Paschen, Brackett and Pfund are infrared.
- Energy-wavelength bridge: ΔE(eV) = 1240/λ(nm); the 2 → 1 jump of 10.2 eV means λ ≈ 121.6 nm, the Lyman alpha line.
- Ionisation numbers: 13.6 eV from the ground state, 3.4 eV from n = 2, 1.51 eV from n = 3 — asked directly as "ionisation potential".
- de Broglie closure: the allowed orbits satisfy 2πr_n = nλ, standing electron waves — the justification Bohr himself lacked, shown in NCERT's chapter.
One numerical, two routes
Ask for the wavelength of the 4 → 2 transition. Route one, Rydberg: 1/λ = 1.097 × 10^7 × (1/4 − 1/16) = 1.097 × 10^7 × 3/16, giving λ ≈ 486 nm, the blue-green Balmer beta line. Route two, energies: ΔE = 13.6 × (1/4 − 1/16) = 2.55 eV, then λ = 1240/2.55 = 486 nm — the same answer from a different toolbox, a redundancy worth practising because the options are built to punish arithmetic slips, not concept. Now invert the direction: what excites hydrogen from n = 1 to n = 3? The atom must absorb 13.6 − 1.51 = 12.09 eV, and a photon of any smaller energy is simply not absorbed at all — discrete rungs mean no partial climbs. That single sentence answers an entire family of NEET questions about which photons a hydrogen atom can absorb.
Where NEET sets the trap
The scaling powers are the trap: radius grows as n², speed falls as 1/n, energy flattens as 1/n² — a question asking "how do r and v change from n = 1 to n = 2" mixes all three, and every wrongly paired option is offered. Negative energies are mined for confusion: the fourth orbit's energy is −0.85 eV, not +0.85 eV, and both appear. Series confusion — asking which series lies in the visible (Balmer alone) or which transition gives the shortest wavelength in a series (the series limit, n2 → ∞) — is standard. One more: "energy required to remove the electron from n = 2" is 3.4 eV, the binding energy, while "energy of the electron in n = 2" is −3.4 eV; the sign carries the meaning, and the options carry both.
Frequently asked questions
What are the radius and energy of hydrogen's n = 2 orbit?
r = 0.529 × 2² ≈ 2.12 Å and E = −13.6/4 = −3.4 eV; ionising the atom from here needs just 3.4 eV more.
Which transition produces the H-alpha line, and at what wavelength?
n = 3 to n = 2, at about 656 nm — the red line that opens the Balmer series, the only visible series of hydrogen.
What is the shortest wavelength in the Balmer series?
364.6 nm, the series limit approached as n2 → ∞: 1/λ = R/4 exactly.
How much energy ionises hydrogen from the ground state?
13.6 eV — the binding energy of the first Bohr orbit; the ionisation potential is likewise 13.6 V.
How does the electron's speed vary with orbit number in Bohr's model?
v_n = 2.18 × 10^6/n m/s — it falls as 1/n; in the ground state the electron moves at roughly c/137.