An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm. The net energy absorbed by the atom in this process is n × 10⁻⁴ eV. The value of n is __________. [Assume the atom to be stationary during the absorption and emission process] (Take h=6.6 × 10⁻³⁴ Js and c=3 × 10⁸ m / s )

Solution

Physics sir · 4 steps

Energy in minus energy out: subtract the two photon energies, then convert to eV.

1. Photon energy. A photon of wavelength λ carries energy E = hc/λ. The atom gains E₁ from the absorbed photon and loses E₂ to the emitted one. Because the atom is stationary, no energy goes into recoil.

   Δ E = hc(1/λ₁-1/λ₂)

2. Put in numbers. hc = 6.6 × 10⁻³⁴ × 3 × 10⁸ = 1.98 × 10⁻²⁵ J m. So E₁ = (1.98 × 10⁻²⁵)/(500 × 10⁻⁹) = 3.96 × 10⁻¹⁹ J and E₂ = (1.98 × 10⁻²⁵)/(600 × 10⁻⁹) = 3.3 × 10⁻¹⁹ J.
3. Convert to eV. Δ E = 3.96 × 10⁻¹⁹ - 3.3 × 10⁻¹⁹ = 0.66 × 10⁻¹⁹ J. Dividing by 1.6 × 10⁻¹⁹ J per eV gives Δ E = 0.4125 eV = 4125 × 10⁻⁴ eV.
4. So. The atom keeps a net 0.4125 eV, so n = 4125.

The idea to keep. The net energy an atom keeps is the absorbed photon energy minus the emitted one. When the question gives h and c, use those values, not the 1240 eV nm shortcut.

Answer from the NTA final answer key. Solution by PrepWiser.