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an electron in a hydrogen like species is in the excited state (n2). the wavelengths corresponding to a transition second orbit is 48.24nm and from same orbit, wavelength corresponding to a transition to third orbit is 142.46 nm. find n2 and Z(atomic number) |
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Answer» Answer: n2 = 5 and Z = 3 Explanation: Here, It is given that an electron in a hydrogen like species is in the excited STATE (n2). the wavelengths corresponding to a TRANSITION second orbit is 48.24nm ∴ n1 = 2 then λ = 48.24 nm => from same orbit, WAVELENGTH corresponding to a transition to third orbit is 142.46 nm. ∴ n2 = 3 then λ = 142.46 nm => To calculate n2 and Z we can use the below formula: 1/λ = R_H*Z² [1/n₁² - 1/n₂²] 1.48.24 = R_H*Z² R_H*Z² [1/n₁² - 1/n₂²]...(1) 1/142.46=R_H*Z² [1/9 - 1/n₂²] ...(2) => DIVIDING eq (1) by (2), we get 142.46/48.24 = n₂² - 4/ n₂² - 9 * 9/4 1.31 = n₂² - 4/ n₂² - 9 n₂ = 5 => By putting this value in equation (1) we can calculate the ATOMIC number: Rh = 109677 cm⁻¹ λ = 48.24 * 10⁻⁷ cm 1/48.24 * 10⁻⁷ = 109677 * Z² [1/4 - 1/25] Z = 3 |
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