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Obtain an epressionfor thefrequencyof radiationemitted whena hydrogen atom de- excites from levle(n-1). Forlarge n, show thatthis frequency equals to classical frequency of revolutionof the electron in the orbit . |
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Answer» Solution :We know that energy of an electron in hydrogenatom in in level is given by `E_(n) = ( me^(4))/(8 in_(0)^(2) n^(2) h^(2))` `THEREFORE `Frequency of radiation emiited when a hydrogen ATOM de - excits from level n to level (n -1) . ` v = (1)/(h) [E_(n) - E_(n-1)] = (m e^(4))/( 8in_(0)^(2) h^(3))[(1)/((n-1)) - (1)/(n^(2))] = (m e^(4))/(8 in_(0)^(2) h^(3)). ((2n-1))/(n^(2) (n-1)^(2))` If n is very LARGE , then `1 lt lt n`andhence , the aboverelationis modified as . `v = (m e^(4))/(4 in_(0)^(2) h^(2))` As perclassical model of atom , linear SPEED of electron in an ORBITOF radius r is given by . ` v= (e)/(sqrt(4 pi in_(0) mr))` andfrequency of radiation = frequencyof orbitat motionof electron . `v_(0) = (v)/(2pir) = sqrt((e^(2))/(16pi^(2) in_(0) mr^(3)))` If we calculatethe value of v and `v_(0)`thenwe findthat theanswerare almostthe same . |
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