An electron in an atom makes a transition from state n=4 to n=2 , emitting a photon of wavelength 480 nm . It then subsequently transitions from n=2 to n=1 , emitting a second photon of wavelength 160 nm . The wavelength of the photon emitted if the electron were to make a direct transition from n=4 to n=1 is

Step-by-step solution

Let the energy levels be E₁, E₂, E₄ . The energy of the photon emitted during the transition from n=4 to n=2 is E₄₂ = E₄ - E₂ = hc ₄₂ . The energy of the photon emitted during the transition from n=2 to n=1 is E₂₁ = E₂ - E₁ = hc ₂₁ . For a direct transition from n=4 to n=1 , the energy is E₄₁ = E₄ - E₁ . Clearly, E₄₁ = E₄₂ + E₂₁ . Substituting the energies in terms of wavelengths: hc ₄₁ = hc ₄₂ + hc ₂₁ 1 ₄₁ = 1 ₄₂ + 1 ₂₁ Given ₄₂ = 480 nm and ₂₁ = 160 nm : 1 ₄₁ = 1 480 + 1 160 = 1 + 3 480 = 4 480 = 1 120 nm ⁻¹ ₄₁ =