Saved Bookmarks
| 1. |
Applying Bohr's theory, find the orbital velocity of the electron on an arbitrary energy level. Compare the orbital velocity on the lowest energy level with that of light. |
|
Answer» `(mv^(2))/r=e^(2)/(4piepsi_(0)r^(2))`, which gives `mv^(2)r=e^(2)/(4piepsi_(0))` The second equation stems from the rule of orbit quantization : mvr = NH. Dividing the first equation by the second, we obtain `v=1/n*e^(2)/(4piepsi_(0)h)` The maximum speed CORRESPONDS to the first (principal) energy level. Its ratio to the speed of light in a vacuum is the find structure constant : `alpha=v_(1)/c=e^(2)/(4piepsi_(0)ch)=7.3xx10^(-3)=1/137` |
|