Ionisation potential of hydrogen atom is 13.6 eV . When hydrogen atoms in ground state are excited by a supply of 12.1 eV , then the number of spectral lines emitted by hydrogen atoms according to Bohr's theory is

Step-by-step solution

Given that, ionisation potential of hydrogen atom, E₁=-13.6 eV Energy supplied = Energy absorbed by hydrogen atom. E=12.1 eV i.e. final energy of excited state, E₂= -13.6 eV n^2 We know that, E₂-E₁= E E₂= E+E₁ Substituting the value, we get aligned & - 13.6 n^2 =[12.1+(-13.6)] & - 13.6 n^2 =-1.50 x^2=9.07 n 3 aligned Hence, by absorbing 12.1 eV energy electron jump into 2nd excited state (n=3) . Then, number of spectral lines emitted by H -atom = n(n-1) 2 = 3(3-1) 2 =3