In a H-atom the energy in the nth orbit is given as En = -13.6n2 ev . Show that En+1 - En = -13.6*2n3 for large n values
How to solve??
11since 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 :)
Anirban Mondal HOW?
PLEASE EXPLAIN THIS
Anik Chatterjee couldnt get ,,....if n=n+1....En+1-En=En-En=0....!!11n 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
