If the average life time of an excited state of hydrogen is of the order of `10^(-8) s`, estimate how many whits an alectron makes when it is in the state `n = 2` and before it suffers a transition to state` n = 1 (Bohrredius a_(0) = 5.3 xx 10^(-11)m)`?

Answer

`8xx10^(6)`

Explanation

`V_(2)=V_(0)xx(1)/(2)=(V_(0))/(2)`
`x=vxxt`
`x=(V_(0))/(2)xx10^(-8)sec=((V_(0)xx10^(-8))/(2))m`
`2pir rarr 1` round
`(V_(0)xx10^(-8))/(2)=(V_(0)xx10^(-8))/(2)xx(1)/(2pir)`
`r_(2)=r_(0)xxn^(2)=4r_(0)`
so, no. of revolutions
`=(V_(0)//2xx10^(-8))/(2pixx4r_(0))=(V_(0)xx10^(-8)xx1)/(2xx2pixx4r_(0))`
`=(2.18xx10^(6)xx10^(-18))/(2xx2xx3.14xx4xx0.529)`
`(2.18xx10^(-12))/(2.6xx10^(-21))=0.838xx10^(9)=8xx10^(6)`