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Atoms - De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Bohr's second postulate states that the angular momentum of an electron orbiting the nucleus is quantized and can only take values that are integral multiples of h2π\frac{h}{2\pi}.

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Louis de Broglie provided a theoretical basis for this postulate by proposing that electrons in atoms behave like standing waves (stationary waves).

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For an electron wave to be stable in a circular orbit of radius rr, the circumference of the orbit must be an integral multiple of its de Broglie wavelength λ\lambda.

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If the circumference 2πr2\pi r is not equal to nλn\lambda, the wave would interfere with itself upon successive revolutions and eventually average out to zero.

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The condition for a standing wave is 2πr=nλ2\pi r = n\lambda, where n=1,2,3,…n = 1, 2, 3, \dots represents the principal quantum number.

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Substituting the de Broglie relation λ=hmv\lambda = \frac{h}{mv} into the standing wave condition yields the quantization of angular momentum: mvr=nh2πmvr = \frac{nh}{2\pi}.

📐Formulae

λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}

2πrn=nλ2\pi r_n = n\lambda

L=mvrn=n(h2π)L = mvr_n = n\left(\frac{h}{2\pi}\right)

n=1,2,3,…n = 1, 2, 3, \dots

💡Examples

Problem 1:

Determine the number of de Broglie wavelengths associated with an electron revolving in the 3rd3^{rd} orbit of a hydrogen atom.

Solution:

According to de Broglie's explanation of Bohr's second postulate, the condition for a stable orbit is given by: 2πrn=nλ2\pi r_n = n\lambda For the 3rd3^{rd} orbit, the principal quantum number is n=3n = 3. Substituting n=3n = 3 into the equation: 2πr3=3λ2\pi r_3 = 3\lambda Therefore, there are 33 de Broglie wavelengths in the 3rd3^{rd} orbit.

Explanation:

In any nthn^{th} orbit, the number of complete de Broglie wavelengths that fit into the circumference of the orbit is exactly equal to the principal quantum number nn.

Problem 2:

Show how the de Broglie hypothesis leads to the quantization of angular momentum for an electron in a circular orbit of radius rr.

Solution:

  1. Start with the standing wave condition for a circular orbit: 2πr=nλ2\pi r = n\lambda
  2. Use the de Broglie relation for wavelength: λ=hmv\lambda = \frac{h}{mv}
  3. Substitute the expression for λ\lambda into the first equation: 2πr=n(hmv)2\pi r = n \left( \frac{h}{mv} \right)
  4. Rearrange the terms to isolate angular momentum (L=mvrL = mvr): mvr=nh2πmvr = \frac{nh}{2\pi}
  5. This is Bohr's quantization condition, where L=nℏL = n\hbar (with ℏ=h2π\hbar = \frac{h}{2\pi}).

Explanation:

By treating the electron as a wave, the requirement for constructive interference (standing wave) naturally results in the restriction of angular momentum to discrete values.