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Question

A photon of energy 10.2 eV is absorbed by a hydrogen atom in the ground state. To which energy level
does the electron jump? (Use \( E_n = -\frac{13.6}{n^2} \, \text{eV} \))

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Explanation

**Bohr's stationary orbits** defined by angular momentum quantization L = n ħ, ħ = h/2π, n=1 ground state, electron in these orbits does not radiate despite acceleration, contrary to classical EM theory which predicts atom collapse due to energy loss via radiation, Bohr postulates to explain observed stability and discrete spectra. E₁ = -13.6 eV . E_n = E₁ + 10.2 = -13.6 + 10.2 = -3.4 eV . -3.4 = -(13.6/n²) ⇒ n² = 4 ⇒ n = 2 . 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 u

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