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(a) Using Bohr's second postulate of quantisation of orbital angular momentum, show that the circumference of the electron in the nth orbital state in hydrogen atom is a times the de Broglie wavelength associated with it. (b) The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally move to the ground state ? |
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Answer» Solution :(a) As per Bohr.s second postulate the angular momentum of orbiting electron is an integer multiple of`(h)/(2pi) ` i.E.,`m v r = n(h)/(2pi) `, where n = 1,2,3,…… `rArr"" 2 pi r = (n h)/(m v )` But`(h)/(m v) = lambda_(e) = de -` Broglie wavelength associated with electron and ` 2 pi r ` = circumferenceof nth electron orbit. Thus, it isprovedthat circumeferene of nth electron orbitis n TIMES the de - Broglie wavelenghtassoicatedwith it . (b) Asthe electron state ,it isin n = 4orbit. Thus, itis provedthatcircumference of nth electron orbits is n times the de - Broglie wavelenghtassociated with it . (b) As the electron is in the third excited state ,it is in n = 4 orbit. Hence, maximum number ofspectral LINES whichbe emittedbefore it comes to groundstate ` = (1)/(2) n(n-1) = (1)/(2)xx 4XX 3 = 6` . |
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