Answer:1870nm
Explanation:
The energy of the hydrogen atom in the quantum state n is E = - (13.6 eV) / n^2. The ground state, when n = 1 and E = - 13.6 eV.
after absorption,the energy of atom is -13.6 + 12.75 = -0.85 eV.
quantum number for this energy state is
-0.85 = -13.6/n^2
n^2 = 13.6/0.85 = 16
n =4.
The next-lower energy orbit has energy
E = -13.6/3^2 = -1.511
Then transition emits a photon of energy -0.85-(-1.511) = 0.661 eV.
you would use
Landa = (planck's constant)(speed of light) / energy
λ = hc/E
or in your case, λ = (6.63 x 10-34 * 3 x 10^8) / (0.661 / 6.24 x 10^18)
to get λ = 1870 nm
