Planck's theory, Bohr's model and the hydrogen spectrum

Chemistry · Atomic structure. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Planck said that energy is not given out or absorbed continuously. It comes in small packets called quanta, and the energy of one quantum (a photon, for light) depends on the frequency of the radiation.

Bohr used this idea for the hydrogen atom. The electron moves only in certain fixed circular orbits, each with a definite energy. It does not radiate while it stays in one orbit. When it jumps from a higher orbit to a lower one, it emits one photon whose energy equals the energy difference.

This explains the line spectrum of hydrogen: each line is one kind of jump. A continuous spectrum, such as a rainbow, contains all wavelengths with no gaps.

Rules to remember

energy of a photon = Planck's constant × frequency (E = hν). Since c = νλ, E = hc/λ. h = 6.626 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹.
Higher frequency means shorter wavelength and more energy. Wave number = 1/λ, in m⁻¹.
Bohr's quantum condition: angular momentum of the electron, mvr = nh/2π, where n = 1, 2, 3 ...
radius of the nth orbit of hydrogen: r = 0.529 × n² Å.
energy of the nth orbit: E_n = −1312/n² kJ mol⁻¹. It is negative because the electron is bound; it is zero at n = ∞. Energy of a jump: ΔE = E_higher − E_lower = hν.
Spectral series, named by the level where the jump ends: Lyman n = 1 (ultraviolet), Balmer n = 2 (visible), Paschen n = 3, Brackett n = 4, Pfund n = 5 (all infrared).
X-rays form when high-speed cathode rays strike a heavy metal target. Moseley's law: √ν = a(Z − b), so X-ray frequency depends on atomic number Z.

Common mistake

Students divide or multiply by n instead of n², or drop the minus sign in E_n. Radius goes up as n²; energy is −1312 divided by n² and becomes less negative as n increases.

Worked example

Using E_n = −1312/n² kJ mol⁻¹, how much energy is emitted per mole when electrons fall from n = 3 to n = 2 in hydrogen, and in which series does the line lie?

E₃ = −1312 ÷ 9 = −145.8 kJ mol⁻¹.
E₂ = −1312 ÷ 4 = −328.0 kJ mol⁻¹.
ΔE = E₃ − E₂ = −145.8 − (−328.0) = 182.2 kJ mol⁻¹.
The jump ends at n = 2, so the line is in the Balmer series.

Answer: 182.2 kJ mol⁻¹ is emitted; the line is in the Balmer series (visible region).

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Chapter 5: Atomic Structure

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