Skip to content

Question

In Bohr’s model, what happens to the energy required to ionize a hydrogen atom as the electron’s orbit
number increases?

Options

Choose one · Correct answer highlighted

Explanation

**Bohr energy levels** E_n = -13.6/n² eV for hydrogen, negative indicating bound state, total energy = -13.6 eV ground state n=1, -3.4 eV n=2, -1.51 eV n=3, etc., photon energy for transition n_i → n_f is ΔE =13.6(1/n_f² -1/n_i²) eV, wavelength λ = hc/ΔE, h=6.6×10⁻³⁴ J·s, c=3×10⁸ m/s. Emission line spectrum characterized by discrete wavelengths because energy levels discrete. As n increases, the energy becomes less negative (closer to zero), so less energy is required to ionize the atom from higher orbits. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.