According to Bohr's theory, Energy of electron in n^{th} orbital of H-atom with Atomic number (Z) = \(-13.6\frac{2^2}{n^2}\)eV

\(\therefore\) The energy of ground state electron in H-atom = -13.6 eV

Now, it absorbs 12.1 eV energy.

So, its energy increase to -13.6 + 12.1 = - 0.85 eV

\(\therefore\) E = -13.6\(\frac{z^2}{n^2}\)eV

-0.85 = -13.6 x \(\frac{(1)^2}{n^2}\) (\(\because\) for hydrogen Z = 1)

\(\therefore\) n^{2 }= \(\frac{13.6}{0.85}\) = 16

\(\therefore\) n = 4

**Hence, the electron in H-atom jumps to 4th orbit on absorbing 12.1 eV energy.**