In a hydrogen atom, the energy levels are given by the formula:E_n = -13.6 eV/n^2where E_n is the energy of the nth level and n is the principal quantum number.For the first excited state n = 2 and the ground state n = 1 , we can calculate the energy difference E between these two levels:E = E_1 - E_2 = -13.6 eV/1^2 - -13.6 eV/2^2 = -13.6 eV + 3.4 eV = -10.2 eVThe energy difference between the first excited state and the ground state is -10.2 eV. The negative sign indicates that energy is released when the electron transitions from the first excited state to the ground state.To find the wavelength of the emitted photon, we can use the equation:E = h * c / where h is Planck's constant 6.626 x 10^-34 Js , c is the speed of light 3 x 10^8 m/s , and is the wavelength.Rearranging the equation to solve for , we get: = h * c / EFirst, we need to convert the energy difference from electron volts eV to joules J using the conversion factor 1 eV = 1.602 x 10^-19 J:E = -10.2 eV * 1.602 x 10^-19 J/eV = -1.634 x 10^-18 JNow, we can calculate the wavelength: = 6.626 x 10^-34 Js * 3 x 10^8 m/s / -1.634 x 10^-18 J 1.215 x 10^-7 mThe wavelength of the transition from the first excited state to the ground state of a hydrogen atom is approximately 1.215 x 10^-7 m or 121.5 nm. This falls in the ultraviolet range of the electromagnetic spectrum.