Energy of a photon: $E = h\nu = \dfrac{hc}{\lambda}$
Energy is inversely proportional to wavelength: $E \propto \dfrac{1}{\lambda}$
The radiation with shorter wavelength (2000 Å) has 3 times more energy than the radiation with longer wavelength (6000 Å).
Planck's equation: $E = h\nu = \dfrac{hc}{\lambda}$, where $h = 6.626\times10^{-34}$ J·s (Planck's constant), $c = 3\times10^8$ m/s (speed of light), $\nu$ = frequency (Hz), $\lambda$ = wavelength (m).
Key relationship: Since $c = \nu\lambda$, we have $\nu = c/\lambda$. Substituting: $E = hc/\lambda$. So energy is inversely proportional to wavelength — shorter wavelength means higher energy photon.
Electromagnetic spectrum in order of increasing energy: Radio < Microwave < Infrared < Visible < UV < X-ray < Gamma rays. Correspondingly, wavelength decreases in this order.
Visible light wavelengths: 400 nm (violet) to 700 nm (red). The two wavelengths here: 2000 Å = 200 nm (UV region) and 6000 Å = 600 nm (orange-red visible). UV photons are 3× more energetic.
Angstrom to nm: 1 Å = 10⁻¹⁰ m = 0.1 nm. So 2000 Å = 200 nm and 6000 Å = 600 nm. Always confirm units before comparing.
Energy of a photon E=hc/λ. Here h and c are constants. For ratio E₁/E₂=λ₂/λ₁=6000/2000=3. Short wavelength radiation is more energetic.
2000 Å = 200 nm falls in the UV range. 6000 Å = 600 nm is orange-red visible light. UV photons have energy ~6.2 eV while orange photons have ~2.1 eV — ratio ≈3. UV can break chemical bonds; visible light generally cannot.
E∝1/λ. If λ doubles, E halves. If λ is 3× larger, E is 3× smaller. Here λ₂=3λ₁, so E₁=3E₂.
1 Å = 10⁻¹⁰ m. 2000 Å = 2×10⁻⁷ m. 6000 Å = 6×10⁻⁷ m. For ratio calculation, units cancel as long as both are in the same unit.