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A hydrogen sample is prepared in a particular excited state A of quantum number, n_A = 3 . The ground state energy of hydrogen atom is - |E| . The photons of energy(|E|)/(12) are absorbed in the sample which results in the excitation of some electrons to excited state B of quantum number n_Bwhose value is |
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Answer» 6 energy of HYDROGEN atom in GROUND state =- | E |and `n_A = 3` `therefore` Energy of electrons in excited state , `n_A` `(E )_(n_A) = (-|E|)/(n_A^2)` Energy of electron in excited state , `n_B` ` = (E )_(n_B) = (-(E ) )/(n_B^2)` When sample ABSORBS the photon of energy, ` (|E|)/(12)`, then its electrons REACHES from energy state `n_A ` to energy state `n_B` Hence , `(E )_(n_B) - (E )_(n_A) = (|E|)/(12)` `(-|E|)/(n_B^2) - { (-|E|)/(n_A^2) } = (|E|)/(12)` `(-1)/(n_B^2) + (1)/(n_A^2) = 1/12 rArr - (1)/(n_B^2) = 1/12 - (1)/(n_A^2)` ` =1/12 - 1/9 rArr - (1)/(n_B^2) = 1/12 - (1)/(n_A^2)` ` = 1/12 - 1/9 rArr - (1)/(n_B^2) = - 1/36` `rArr n_B^2 = 36 ` ` therefore n_B = 6` |
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