Question 32 of 62advanced🔍 AnalyzeLong Answer3 marks

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Correct Answer

According to Bohr's model, the angular momentum of an electron revolving in a circular orbit is quantized and is given by mvr=nh2πmvr = \frac{nh}{2\pi}, where nn is an integer (principal quantum number). According to de Broglie, the wavelength associated with a moving electron is λ=hmv\lambda = \frac{h}{mv}.

From the de Broglie equation, we can write mv=hλmv = \frac{h}{\lambda}. Substituting this value of mvmv into Bohr's quantization condition, we get: hλ×r=nh2π\frac{h}{\lambda} \times r = \frac{nh}{2\pi}. Rearranging this equation gives 2πr=nλ2\pi r = n\lambda.

Here, 2πr2\pi r represents the circumference of the Bohr orbit. The equation 2πr=nλ2\pi r = n\lambda clearly shows that the circumference of the orbit is equal to nn times the de Broglie wavelength. Thus, the circumference is an integral multiple of the de Broglie wavelength.

Exercise: EXERCISES | Q: 2.32 | (Page 41)
For More Understanding

Explanation

The derivation links Bohr's quantization of angular momentum with de Broglie's matter wave hypothesis. By substituting the momentum term from the de Broglie equation into Bohr's angular momentum equation, we mathematically prove that an electron's orbit must perfectly fit a whole number of wavelengths, justifying the stable orbits in Bohr's model.

Solution Steps

  1. Step 1: Write Bohr's quantization condition: mvr=nh2πmvr = \frac{nh}{2\pi}

  2. Step 2: Write de Broglie wavelength equation: λ=hmv\lambda = \frac{h}{mv}, which gives mv=hλmv = \frac{h}{\lambda}

  3. Step 3: Substitute mvmv into Bohr's condition: hλr=nh2π\frac{h}{\lambda} r = \frac{nh}{2\pi}

  4. Step 4: Rearrange to find circumference: 2πr=nλ2\pi r = n\lambda, proving it is an integral multiple of λ\lambda.

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