at Calculate the frequency (Hz) of the emitted photon when an electron drops from the n = 4 to the n = 2 level in a * hydrogen atom

QUESTION:

Firstly-

• N4-n2=E4-E2=-o,85-(-3,39) = 2,54 eV

So the electron has an energy of 2,54 eV=4.064e-19

Stape 2-

E=hxc/wavelength  so wavelength=hxc/E

Wavelength= 6,34e-34 x 3e8/4,064e-19=4,68 e-7

So 468 nm

Stape 3-

• Wavelength = c x T <-> T= wavelength/c

T=468e-9/3e8=1,54 e-15

Stape 4-

f=1/T = 6,4 e14 Hz