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⚛️ NEET Physics❓ Question

de Broglie explaining Bohr's angular momentum rule, plus what exactly are the limitations of the Bohr model?

Mehak AroraDropper Sample

10 Sept 2026 · 1 min read

Sorry for posting at 1 am, can't sleep and this chapter is stuck in my head.

NCERT says de Broglie's hypothesis explains Bohr's second postulate. I followed the algebra: if the orbit is a standing wave then

2πrn=nλ,λ=hmv⇒mvrn=nh2π2\pi r_n = n\lambda, \quad \lambda = \frac{h}{mv} \Rightarrow mvr_n = \frac{nh}{2\pi}

OK. But why must the circumference be a whole number of wavelengths? I know standing waves on a string need nodes at the ends, but a circle has no ends. And historically did Bohr know about this? Because Bohr's model came first, I think?

Then the second part: "limitations of Bohr model". In my notes I only have 'works only for hydrogen'. Last year's board paper wanted more and I lost marks. What are the actual limitations that NEET could ask? I also read that it violates the uncertainty principle but I don't see how, since the orbit radius and the momentum are both 'known' in Bohr's picture.

Please explain slowly, my brain is tired. 🙂

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6 answers & replies

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  • [+]Nisha Verma12thSample· 1mo ago

    Whole number because otherwise the wave doesn't join up with itself. Go around the circle once: if the wave ends at a different phase than where it started, the next lap is out of phase with the first, and after many laps the waves cancel out (destructive interference). Only when 2πr=nλ2\pi r = n\lambda does each lap overlap exactly on the last one, giving a stable standing wave. A string doesn't need ends for that, the 'joining up' condition replaces the fixed ends.

    And yes, Bohr's postulate (1913) came way before de Broglie (1924). de Broglie gave the reason after the fact.

  • [+]Aman YadavDropperSample· 1mo ago

    Limitations, short list. Learn this.

    1. Only for one-electron species: H, He⁺, Li²⁺.
    2. Cannot explain fine structure or intensity of spectral lines.
    3. Cannot explain splitting in magnetic field (Zeeman) or electric field (Stark).
    4. Does not explain why only certain orbits are stable.
    5. Violates the uncertainty principle.
    6. Cannot explain molecule formation / chemical bonding.

    Done.

  • [+]Harsh Gupta12thSample· 1mo ago

    Bohr also works for sodium na, since sodium has just one valence electron so you can treat it like hydrogen with a big nucleus. The inner electrons just make the nucleus look a bit different, no? Also I learnt this from a YouTube video at 2x speed, so I'm basically a professor. Ok I'll show myself out 😅

    • [+]Arjun Nair12thSample· 1mo ago

      Harsh, no. Sodium has 11 electrons and the inner ones screen the nucleus, and the valence electron also interacts with them, so the simple 1/n21/n^2 energy formula fails. Bohr's formulas are valid only for hydrogen-like ions, meaning a nucleus plus exactly one electron.

      Mehak, on uncertainty: Bohr's orbit says the electron has a definite radius AND definite momentum at the same time. Heisenberg says Δx Δp≳h/4π\Delta x\,\Delta p \gtrsim h/4\pi, so you can't have both exactly. That's the violation.

      • [+]Mehak AroraDropperSample· 1mo agoAuthor

        thank you both, the Zeeman and Stark ones I had never even seen. Will rewrite my list (and not trust Harsh for chemistry 😂)

  • [+]Tushar PawarDropperSample· 1mo ago

    A quick numerical to feel it. In the first orbit of H, v=2.19×106v = 2.19\times10^6 m/s, so λ=hmv=6.63×10−349.1×10−31×2.19×106≈3.3×10−10 m\lambda = \frac{h}{mv} = \frac{6.63\times10^{-34}}{9.1\times10^{-31}\times 2.19\times10^{6}} \approx 3.3\times10^{-10}\text{ m} and the circumference is 2π×0.5292\pi \times 0.529 Å =3.32= 3.32 Å. Same number, exactly one wavelength fits. In the n-th orbit λn=2πrn/n∝n\lambda_n = 2\pi r_n/n \propto n (since r∝n2r \propto n^2). That ratio shows up in MCQs a lot.

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