AIPMT 2014 Q41 Please provide a detailed solution.......

energy of a photon = E= hc / lambda = 12.75 ev 
the enrgy is equal to energy gap between n = 1 (-13.6 ) and n = 4 (-0.85 ) so, by this energy the electron will exite from n= 1 to n= 4 . , 
when the electron will fall back , the no of spectral lines emitted = n (n-1 ) / 2 
4(4-1) / 2= 6 
its the answer , 
its acc to the question paper that i have , but dnt know its your ques or not as u have not attached any ques . 
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