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A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted ? |
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Answer» Solution :We know that energy of an electron in nth orbit of hydrogen is `E_(n)= -(13.6)/(n^(2))` and energy in ground STATE of hydrogen is -13.6 eV. Whenan electron beam of 12.5 eVis used , hydrogen atom cannot be ionised but atthe mostexcitedto n= 3state where energy of theelectron will be ` - (13.6)/((3)^(2)) = - 1.51 eV`(as maximum energy of electron can be`- 13.6 + 12.5 = - 1.1eV` andelectron cannot be excited to n = 4state having energy - 0.85 eV) So only THREE transitions are possible So only three transitions are possibleasshwonin Fig. 12.01. Here transtionsNo.1 and3 represent lins of Lyman series . andtransition No.2 represents lineof Balmer series . Wavelength of lightemiited in a transition will be . `lambda = (c)/(v) = (ch)/((E_(i) -E_(f))` ` therefore""lambda_(1) = (3 xx10^(8) xx 6.63 xx 10^(-34))/([(-1.51 - (- 13.6)] xx 1.6 xx 10^(-19)))= (3 xx 10^(8) xx 6.63 xx 10^(-34))/(12.09 xx 1.60 xx 10^(-19)) = 1.03 xx 10^(-7) m = 103 nm` `lambda_(2) = (3 xx 10 xx 6.63 xx 10^(-34))/([(-1.51 - (-3.4)] xx1.6 xx 10)^(-19)) = (3 xx10^(8) xx 6.63 xx 10^(-34))/(1.89 xx 1.6 xx 10^(-19)) = 6.56 xx 10^(-7) m = 656 nm` and`lambda _(3) =(3 xx 10^(8) xx 6.63 xx 10^(-34))/([(-3.4- (-13.6)]xx1.6 xx 10^(-19)))= (3 xx 10^(8) xx 6.63 xx 10^(-34))/(10.2 xx 1.6 xx 10^(-19)) = 1.22 xx 10^(-7) m =122 nm` 
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