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)`?

If the average life time of an excited state of hydrogen is of the ord

1.	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» Correct Answer - `8xx10^(6)` `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)`

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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)`?

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