Atomic Structure

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In a H-atom the energy in the nth orbit is given as E_n = -13.6n² ev . Show that E_n+1 - E_n = -13.6*2^n³ for large n values

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Debopam Purkayasthm ·Jun 15 '12 at 7:14

since n is infinitely large,n=n+1(approx)....again, 1/n is negligible....so take it 0...nd u will be able 2 solve d problm :)

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Anirban Mondal HOW? PLEASE EXPLAIN THIS
Upvote·0· Reply ·Jun 16 '12 at 23:30
Anik Chatterjee couldnt get ,,....if n=n+1....En+1-En=En-En=0....!!
Upvote·0· Reply ·Jun 17 '12 at 2:44
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Debopam Purkayasthm ·Jun 30 '12 at 20:56

n approximtely equal to n+1....so the equatn will be -13.6[1/(n+1)² - 1/n²]= -13.6{-2n-1/n²(n+1)²}= 13.6{2n+1/n²(n+1)²}=13.6{2n/n²(n+1)²+1/n²(n+1)²}...now since n is infinitely lrge...n=n+1 approx...again...1/n⁴ is negligible.....so,the equatn stnds lyk 13.6{2n/n⁴ +1/n⁴},or,13.6{2/n³ +0}...which is equal to 13.6 *2/n³

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Atomic Structure

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