Light shines through atomic hydrogen gas. It is seen that the gas absorbs light readily at a wavelength of 91.63 nm. What is the value of n of the level to which the hydrogen is being excited by the absorption of light of this wavelength? Assume that the most of the atoms in the gas are in the lowest level. (h = 6.626 × 10^-34 J ∙ s, c = 3.00 × 10⁸m/s, 1 eV = 1.60 × 10^-19 J, the Rydberg constant is R = 1.097 × 10⁷ m^-1) A) 14 B) 16 C) 11 D) 21

Question

Quizplus · Accepted Answer

The energy of a photon is given by E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The energy of a photon in electron volts is given by E (eV) = hc/λ (eV), where hc/λ (eV) is the energy of a photon in electron volts. The energy difference between two levels in an atom is given by ΔE = E2 - E1, where E2 is the energy of the higher level and E1 is the energy of the lower level. The energy difference between two levels in an atom can be expressed in terms of the Rydberg constant R as ΔE = -Rhc/n2, where n is the principal quantum number of the higher level. Substituting the given values into the equation, we get:ΔE = -Rhc/n2 = -(1.097 × 107 m-1)(6.626 × 10-34 J ∙ s)(3.00 × 108 m/s)/n2 = -2.18 × 10-18 J/n2The energy of a photon in electron volts is given by E (eV) = hc/λ (eV) = (6.626 × 10-34 J ∙ s)(3.00 × 108 m/s)/(91.63 × 10-9 m) = 2.18 × 10-18 JTherefore, the energy difference between the two levels is:ΔE = E2 - E1 = 2.18 × 10-18 JSubstituting this value into the equation ΔE = -Rhc/n2, we get:2.18 × 10-18 J = -(1.097 × 107 m-1)(6.626 × 10-34 J ∙ s)(3.00 × 108 m/s)/n2Solving for n, we get:n = 14

Light Shines Through Atomic Hydrogen Gas