The wavelength in Angstrom units is given by,\lambda =\frac{12,400}{E_{2}-E_{1} }

Since the hydrogen atom goes from n = 10 state to the ground state,\lambda =\frac{12,400}{E_{10}-E_{1} }

The energy of the 10^{th} state is E_{10}=\frac{-13.6}{10^{2} }=-0.136eV.

The energy in the ground state is E_{1}=-13.6eV.

Thus the wavelength of the emitted photon =\frac{12,400}{-0.136-(-13.6)}=920.97 Å