The Bohr Theory - an example

?E = hf = [-13.6 eV]*[(1/nf2) - (1/ni2)]

In the case of ni = 3, and nf = 2,

?E = (-13.6 eV)*(1/4 - 1/9) = 1.89 eV

?E = hf = hc/? , so in this case,

?????emitted = hc/?E =

(6.63x10-34 J-sec)*(3x108 m/s)/(1.89 x 1.6x10-19 J)

= 658 nm (red light).

Previous slide

Next slide

Back to first slide

View graphic version