If the average life time of an excited state of H atom is of order `10^(-8)` sec, estimate how many orbits an `e^(-)` makes when it is in the state `n=2` and before it suffers a transition to `n=1` state.

Answer

`8xx10^(6)`

Explanation

No.of revolutions `=("Total time")/("Time taken in one revolution ")`
`10^(-8)/=((2pir)/V)=(10^(-8)xxV_(0)Z^(2))/(2pir_(0)n^(3))=(10^(-8)xx2.18xx10^(6)xx1^(3))/(2pixx.529xx10^(-10)xx2^(3))=8xx10^(6)`