If 9.9 eV energy is supplied to H atoms the no. Of spectral lines emitted is equal to (a) 0(b) 1(c) 2(d) 3

9.9 eV taken to be approx. as 10 then number of spectral lines  could be =>

n(n-1)/2 = 10

we get,

n^2 – n -20 = 0 

on solving the quadratic we get,

 n = 5.

5 spectral lines are possibly emitted.

Acc to ques energy given to H atom is 9.9ev.  Let’s first calculate energy required to make lowest emmision possible which will be from orbit 2^nd to orbit 1^st so energy diff b/w them will be....13.6(1-1/4) which comes out to be 10.2....since energy given to H atom acc to ques is less as compared to enrgy required to make lowest emmision possible hence no emmision can take place...so answer will be (a) i.e 0