calculate the frequency & wavelength of the emitted photon when electron drops from n=4 to n=2 in hydrogenatom

Question

calculate the frequencey and wavelength (nm) of the emitted photon when an electron drops from the n=4 to the n=2 level in a hydrogen atom?

Answer 1

Without considering the fine structure, the electron energy level of the hydrogen atom can be written as:
En = -2.179×10^-18 J/ n^2
Thus for n = 4: E4 = -1.362x10^-19 J
For n = 2: E2 = -5.448x10^-19 J
Hence the energy of the emitted photon is:
E4 - E2 = 4.086x10^-19 J
To convert energy to frequency or wavelength, we have to use the following formula:
E = hf = hc/L
where h the Plank constant 6.626x10^-34 J*s, f the frequency in 1/s, c the speed of light 2.998x10^8m/s, and L the wavelength in m (1nm = 10^-9m). Now you do the math.