The ionization energy of the hydrogen atom is 13.6 eV.What is the wavelength of a photon having this much energy? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.6 x 10^-19 J,and 1 nm = 10^-9 m) A)91.4 nm B)136 nm C)273 nm D)360 nm

Question

Quizplus · Accepted Answer

The energy of the photon is 13.6 eV, which is equivalent to 13.6 x 1.6 x 10^-19 J. Using the formula E = hc/λ, where h is Planck's constant and c is the speed of light, we can solve for λ (wavelength). λ = hc/E = (6.63 x 10^-34 J·s)(3.00 x 10^8 m/s) / (13.6 x 1.6 x 10^-19 J) = 91.4 nm.

The Ionization Energy of the Hydrogen Atom Is 13