Bohr Jump Lab — a Lyman transition in hydrogen — JEE Advanced 2026 Paper 1 Q.5

01The question

Q.5 Consider a hydrogen atom with v_k, r_k, and K_k denoting the velocity, orbital radius and kinetic energy of the electron in the k^th orbit, respectively. The electron undergoes a transition from the n^th orbit, emitting radiation corresponding to the Lyman series. Considering h to be the Planck’s constant and ϵ₀ the permittivity of the free space, the correct statement(s) is/are:

(A)Magnitude of change in kinetic energy of electron can be expressed as h/4π · |n v_n/r_n − v₁/r₁|.
(B)Magnitude of change in de Broglie wavelength of the electron can be expressed as e²/4ϵ₀ · |1/K_n − 1/K₁|.
(C)Frequency of the radiation emitted can be expressed as e²/(8πϵ₀h) · (1/r₁ − 1/r_n).
(D)Magnitude of change in total energy of the electron can be expressed as h/2π · |v₁/r₁ − n v_n/r_n|.

02The experiment

Answer— Press START: the electron jumps, and each expression is checked against the measured change.

Ready.

From orbit n

Speed

Wave

03Detailed solution

Answer: —

Bohr's rule links everything. In orbit k: m v_k r_k = k h/2π. The Coulomb force supplies the centripetal force: m v_k²/r_k = e²/(4πϵ₀r_k²).
K_k = ½m v_k² = ½(m v_k r_k)(v_k/r_k) = (h/4π) · k v_k/r_k and K_k = e²/(8πϵ₀r_k).
(A) is true. |ΔK| = |K_n − K₁| = (h/4π)|n v_n/r_n − v₁/r₁|.
(D) is false. The total energy is E_k = −K_k, so |ΔE| = |ΔK| = (h/4π)|…|. The expression in (D) has h/2π and is twice the true change.
(C) is true. The photon carries away hν = E_n − E₁ = K₁ − K_n = (e²/8πϵ₀)(1/r₁ − 1/r_n).
ν = e²/(8πϵ₀h) · (1/r₁ − 1/r_n).
(B) is false. λ_k = h/(m v_k) = 2πr_k/k. This is the standing wave the experiment draws: k wavelengths fit around orbit k. With r_k = e²/(8πϵ₀K_k):
λ_k = e²/(4ϵ₀ k K_k) ⇒ |Δλ| = (e²/4ϵ₀)|1/(n K_n) − 1/K₁|.
Option (B) drops the 1/n, so it is wrong for every Lyman line (n ≥ 2). For n = 3 it gives 26.6 Å instead of the true 6.65 Å.

Jump 3 → 1	Measured change	Expression in the option	Verdict
(A) \|ΔK\|	12.09 eV	12.09 eV	True
(B) \|Δλ\|	6.65 Å	26.6 Å	False (missing 1/n)
(C) ν	2.92 × 10¹⁵ Hz	2.92 × 10¹⁵ Hz	True
(D) \|ΔE\|	12.09 eV	24.19 eV	False (twice the change)

Understand the concept

In the Bohr model one rule, m v r = kħ, fixes the whole orbit. The speed falls as 1/k, the radius grows as k², the kinetic energy falls as 1/k², and the de Broglie wavelength grows as k. Exactly k wavelengths wrap around orbit k. The total energy is −K. When the electron drops to orbit 1 it speeds up and its kinetic energy rises, but its total energy falls by the same amount. That energy leaves as a Lyman-series photon in the ultraviolet.

Key takeaways

K_k = (h/4π) k v_k/r_k = e²/(8πϵ₀r_k), and E_k = −K_k.
|ΔE| = |ΔK|: the same expression, not twice it.
λ_k = 2πr_k/k = e²/(4ϵ₀k K_k). Keep the 1/k.
Lyman: hν = K₁ − K_n = 13.6(1 − 1/n²) eV, which is ultraviolet.

Verification: an independent derivation, a sympy route, an exact numerical route with CODATA constants for every Lyman line n = 2…6, and the page's own engine, run by tests/verify_p1phyq05.py. The official key (A, C) and the MathonGo solution (A, C) agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded. Orbit radii are not drawn to scale (the true ratio is k²), and the ultraviolet photon is drawn violet.

How to use this simulation

Choose the starting orbit with From orbit n (2 to 6), then press ▶ START EXPERIMENT.
Watch the electron and its de Broglie wave on orbit n: exactly n wavelengths fit around it.
The electron jumps to orbit 1 and emits a photon. The readouts show v, r, K and λ before and after.
The last stage checks each option's expression against the measured change: ✓ for (A) and (C), ✗ for (B) and (D).
Replay runs it again. Reset clears it. Classroom fills the screen for a projector.