# Bohr Model Numericals

> Radius, energy and transition wavelength numericals from Bohr's model and the Rydberg formula for NEET Physics.

- Canonical URL: https://prepelephant.com/topics/neet-ug/physics/bohr-model-numericals-neet
- Exam / course: NEET-UG · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Bohr Model Numericals", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/bohr-model-numericals-neet

## 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.