de Broglie Explanation of Bohr Quantisation
de Broglie's explanation says that an electron in a stable Bohr orbit behaves as a matter wave forming a standing wave around the nucleus, so the circumference of the orbit must contain an integral number of wavelengths: 2πr = nλ.
In one line: A permitted orbit is a closed wave loop with no loose end.
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If the electron wave does not fit exactly around the orbit, the wave would not join smoothly with itself and the orbit would not be stable. Stable orbits occur only when the circular path length equals a whole number of de Broglie wavelengths. Using λ = h/mv, the condition 2πr = nλ gives mvr = nh/2π, which is Bohr's angular momentum quantisation.
Example
Formula and rule
Standing-wave condition: 2πr = nλ, where r is orbit radius in m, n is a positive integer, and λ is de Broglie wavelength in m. de Broglie relation: λ = h/mv, with h in J s, m in kg, and v in m s−1.
Combining them gives mvr = nh/2π.
See it worked out
Follow the example line by line, then use the same method on new questions.
Worked example
Show that de Broglie's condition gives Bohr's rule.
- 1Start with 2πr = nλ.
- 2Substitute λ = h/mv.
- 3Then 2πr = nh/mv.
- 4Multiplying both sides by mv gives 2πmvr = nh.
- 5Therefore, mvr = nh/2π.
- 6Final result: angular momentum is quantised.
Method you can reuse
- 1State that electron has de Broglie wavelength λ = h/mv.
- 2State stable orbit condition 2πr = nλ.
- 3Substitute λ.
- 4Rearrange carefully to get mvr = nh/2π.
- 5Explain that n must be an integer for a standing wave.
Visual tip
A diagram may show a circular orbit with one, two, or three complete matter wavelengths fitted around it. Label orbit radius r, wavelength λ, and the condition 2πr = nλ.
Check yourself
Answer the question in your head before opening the answer.
Try it yourself
How does de Broglie's idea justify Bohr's quantisation condition?
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Common wrong answer
A common wrong answer is saying quantisation happens because the electron's speed is zero in a stationary orbit. Stationary orbit means fixed allowed energy, not a stationary electron.
Don't mix this up
Exam tip
Write it in the exam
Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.
How CBSE asks it
- 1
Enough for 1 mark
de Broglie's explanation says that an electron in a stable Bohr orbit behaves as a matter wave forming a standing wave around the nucleus, so the circumference of the orbit must contain an integral number of wavelengths: 2πr = nλ.
- 2
Add for 2 marks
Example: For the first Bohr orbit, n = 1, so the circumference contains one complete de Broglie wavelength. For n = 2, it contains two complete wavelengths.
- 3
Add for 3 marks
If the electron wave does not fit exactly around the orbit, the wave would not join smoothly with itself and the orbit would not be stable.
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