Virtual laboratory/Physics/Electromagnetic waves/Photon energy — color and energy

Photon energy — color and energy

Slide the wavelength from radio waves to ultraviolet. The photon energy E=hc/λ grows as the wave shortens — that is why UV and X-rays are dangerous, while radio waves are harmless. Watch the EM wave and the energy bar.

Photon energy — color and energy
Physics · Electromagnetic waves · Photon energy (FIZ.EM.PP.05)
Wavelength λ\lambda (log scale)501 nm
A photon with a shorter wavelength carries energy that is:
wavelength λ
501 nm
frequency f
598.18 THz
energy E
2.47 eV
band
visible light
🎯 Challenge: hit the photon energy
streak: 0

I will give you a target photon energy — pick the wavelength λ to reach it. E = hc/λ = 1240/λ[nm].

What just happened?

You shortened the wavelength — the wave on screen got denser, and the energy packet (the photon) grew. A shorter wave means a higher frequency, and a higher frequency means greater photon energy.

Photon energy:

E=h f=h cλE = h\,f = \dfrac{h\,c}{\lambda}

— the photon energy is directly proportional to the frequency and inversely proportional to the wavelength. hh — the Planck constant, cc — the speed of light, λ\lambda — the wavelength, f=c/λf=c/\lambda — the frequency.

What does it mean in practice?

All EM waves travel at the same speed cc, so a shorter wave "fits" more oscillations per second → higher ff → greater EE. A handy formula in electronvolts:

E [eV]=1240λ [nm]E\,[\text{eV}] = \dfrac{1240}{\lambda\,[\text{nm}]}

Why are UV and X-rays dangerous, but radio is not?

A UV or X-ray photon carries thousands of times more energy than a photon of visible or radio light — enough to break bonds in molecules (ionization). A radio photon is so "light" that it cannot.

Why you need this for the exam:

E=hf=hc/λE = hf = hc/\lambda is the basis of problems on photons, the photoelectric effect and spectra. Trick: compute the frequency f=c/λf=c/\lambda, then the energy E=hfE=hf — or straight from E=1240/λ[nm]E=1240/\lambda[\text{nm}] in eV.

Remember: a shorter wave → higher frequency → greater photon energy. E=hc/λE = hc/\lambda, so E∝1/λE \propto 1/\lambda.

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Theory & formulas

Photon energy

E=hf=hcλE = hf = \dfrac{hc}{\lambda} A shorter wave carries greater energy: a violet photon > a red one, and UV and X-rays > visible light.

Why UV is harmful

A UV photon's energy is enough to break chemical bonds in cells; radio-wave photons are too weak.

Symbols: EE — photon energy, h≈6.63⋅10−34 J sh\approx6.63\cdot10^{-34}\ \mathrm{J\,s} — the Planck constant, ff — frequency, cc — the speed of light, λ\lambda — wavelength.

Frequently asked questions

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Which photon has greater energy?

The one with the shorter wavelength (higher frequency): E=hf=hc/λ, so a violet photon carries more energy than a red one.

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Why is UV more dangerous than visible light?

It has a shorter wave, so a greater photon energy — enough to break chemical bonds in cells.

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What is the Planck constant?

The coefficient h in E=hf that links photon energy to frequency (h≈6.63·10⁻³⁴ J·s).

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