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In Li⁺⁺, electron in first Bohr orbit is excited to a level by a radiation of wavelength λ. When the ion gets deexcited to the ground state in all possible ways(including intermediate emission) a total of six spectral lines are observed. What is the value of λ?(Given: h = 6.63 × 10³⁴ js; e = 3 × 10⁸ ms⁻¹)(A) 10.8 nm (B) 11.4 nm(C) 9.4 nm (D) 12.3 nm |
Answer» answer : OPTION (A) 10.8 nmgiven, total number of SPECTRAL lines are observed = 6 ⇒n(n - 1)/2 = 6 ⇒n² - n - 12 = 0 ⇒n² - 4n + 3n - 12 = 0 ⇒n = 4, - 3 so, n = 4 hence transmission from n₁ = 1 to n₂ = 4 now ENERGY , ∆E = 13.6 × Z²[1/n₁² - 1/n₂² ] = 13.6 × (3)² [1/1² - 1/4²] = 13.6 × 9 × 15/16 = 114.75 eV now using FORMULA, ∆E = hc/λ λ = 1240/∆E (in eV) nm = 1240/114.75 = 10.8 nm hence option (A) is correct choice.. |
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