What is the ionization energy of a hydrogen atom if the electron is in its ground state? E = 1.09678 × 10^-2 nm^-1 R = 1.09678 × 10^-2 nm^-1 Ionization energy = E * R = 1.09678 × 10^-2 nm^-1 * 1.09678 × 10^-2 nm^-1 = 1.2034 × 10^-4 nm^-2 = 1.2034 × 10^-4 kJ/mol
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The formula for the energy of an electron in a hydrogen atom is given by E = -R/n^2, where R is the Rydberg constant and n is the principal quantum number. For the ground state of a hydrogen atom, n = 1. So, E = -R/1^2 = -R. Given that R = 1.09678×10^−2 Show more…
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