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Difficulty: HardAtomic Models

In Bohr's atomic model of the hydrogen atom, the radius of the ground state orbit (n=1n = 1) is 0.053 nm0.053\text{ nm}. According to de Broglie's condition for stationary electron orbits, what is the de Broglie wavelength of the electron in its second excited state? Express your answer in nanometers (nm\text{nm}).

Answer: 1 nm

Answer

The de Broglie wavelength of the electron in its second excited state is 1.00 nm1.00\text{ nm}.
The second excited state corresponds to the quantum number n=3n = 3. According to Bohr's atomic model, the radius of the nn-th orbit is given by rn=n2r1r_n = n^2 r_1, yielding r3=32×0.053 nm=0.477 nmr_3 = 3^2 \times 0.053\text{ nm} = 0.477\text{ nm}. De Broglie explained Bohr's angular momentum quantization by showing that an integral number of electron matter-waves must fit around the orbital circumference: 2πrn=nλn2\pi r_n = n \lambda_n. Solving for λ3\lambda_3 gives λ3=2πr33=2π×3×0.053 nm≈1.00 nm\lambda_3 = \frac{2\pi r_3}{3} = 2\pi \times 3 \times 0.053\text{ nm} \approx 1.00\text{ nm}.

Step-by-Step Solution

1
Identify the principal quantum number for the specified energy state
n=3n = 3
The ground state corresponds to n=1n = 1, the first excited state to n=2n = 2, and the second excited state to n=3n = 3.
2
Calculate the radius of the third stationary orbit
r3=0.477 nmr_3 = 0.477\text{ nm}
In Bohr's model, the radius of the nn-th orbit is proportional to n2n^2, so r3=32×0.053 nm=9×0.053 nm=0.477 nmr_3 = 3^2 \times 0.053\text{ nm} = 9 \times 0.053\text{ nm} = 0.477\text{ nm}.
3
Apply de Broglie's standing wave condition for stationary orbits
λ3=2πr33\lambda_3 = \frac{2\pi r_3}{3}
De Broglie postulated that a stationary orbit contains an integral number of electron de Broglie wavelengths around its circumference: 2πrn=nλn2\pi r_n = n \lambda_n.
4
Substitute values to compute the wavelength
λ3=1.00 nm\lambda_3 = 1.00\text{ nm}
λ3=2×3.1416×0.477 nm3=2π×3×0.053 nm≈0.9992 nm≈1.00 nm\lambda_3 = \frac{2 \times 3.1416 \times 0.477\text{ nm}}{3} = 2\pi \times 3 \times 0.053\text{ nm} \approx 0.9992\text{ nm} \approx 1.00\text{ nm}.

Key Concept

De Broglie Standing Wave Quantization in Bohr Atomic Model
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Topics

Atomic Models

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