Skip to content

Question

A hydrogen atom absorbs a photon of energy 1.89 eV from the \( n = 2 \) state. To which energy level
does it jump? (Use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \))

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. E₂ = -3.4 eV . E_n = -3.4 + 1.89 = -1.51 eV . -1.51 = -(13.6/n²) ⇒ n² = 9 ⇒ n = 3 . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm

Discussion

Comments

0 comments

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