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A hydrogenatom in the groundstate is excited by an electron beam of 12.5 eV energy. Findout themaximum numberof linesemittedby theatom from its excited state. |
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Answer» Solution :We know that energy of anelectron in nth orbit of hydrogen is `E_(n) = - (13.6)/(n^(2)) E V`andenergyis groundstateof HYDROGENIS - 13.6 eV. When an electron beam of 12.5 eVis used, hydrogen atom cannotbe ionisedbut at the mostexcitedto n = 3state whereenergyof theelectron will be ` - (13.6)/((3)^(2)) = - 1.51 eV`(as MAXIMUM energy ofelectron can be` - 13.6 + 12.5 = - 1.1 eV` and electroncannot beexcited to n = 4statehavingenergy- 0.85 e V). So, only three transitions are possibleas show here . Heretransitions No.1and2 represent lines of LYMAN series andtransition No.2 represents lineofBalmer series . 
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