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Chemistry · General chemistry I · Worked example

Compare microwave and visible photons

Find the frequency and photon energy of microwaves with wavelength 12.2 cm, and compare them with the 425 nm light of the first example.

ν=cλ

Find the frequency

12.2 cm = 0.122 m, so ν = c/λ = 2.46 × 10⁹ Hz.

ν=2.998×108 m/s0.122 m=2.46×109 s−1
ν=2.998×108 m/s0.122 m=2.46×109 s−1

Find the photon energy

E = hν = (6.626 × 10⁻³⁴ J·s)(2.46 × 10⁹ s⁻¹) = 1.63 × 10⁻²⁴ J per photon, or about 0.98 J per mole of photons.

E=(6.626×10−34 Js)⁢(2.46×109 s−1)=1.63×10−24 J
E=(6.626×10−34)⁢(2.46×109) J=1.63×10−24 J

Compare

The wavelengths differ by a factor of 0.122/(4.25 × 10⁻⁷) ≈ 2.87 × 10⁵, and so do the photon energies: 281 kJ/mol against about 0.98 J/mol. Microwave photons are far too weak to break chemical bonds; they set polar molecules such as water rotating, which warms the food.

Result

ν = 2.46 × 10⁹ Hz and E = 1.63 × 10⁻²⁴ J per photon, about 2.87 × 10⁵ times less than a 425 nm photon.

Your turn

What is the frequency of radio waves with wavelength 3.00 m?

Show the answer and explanation

9.99 × 10⁷ Hz, about 100 MHz.

ν = c/λ = (2.998 × 10⁸ m/s)/(3.00 m) = 9.99 × 10⁷ s⁻¹.

ν=2.998×108 m/s3.00 m=9.99×107 s−1
ν=2.998×108 m/s3.00 m=9.99×107 s−1

Keep exploring

In Photons & spectrophotometry, make the wavelength ten times longer, 1.22 m. The frequency and the energy per photon both fall tenfold, to 2.46 × 10⁸ Hz and 1.63 × 10⁻²⁵ J.

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