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How many revolutions does an electron in the n = 2 state of a hydrogen atom make before dropping to the n = 1 state ? The average lifetime of an excited state is 10^(-8) s. Bohr radius 0.53 Å and velocity V_(1)=2.19xx10^(6)m//s |
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Answer» Solution :Velocity of electron in the nth orbit of hydrogen atom `v_(n)=(v_(1))/(n)` `=(2.19xx10^(6))/(n)` `:.v_(2)=(2.19xx10^(6))/(2)=1.095xx10^(6)m//s` and radius in `n^(th)` orbit `r_(n)=n^(2)r_(1)` `=(2)^(2)xx0.53Å` `=4xx0.53Å` `=2.12Å` `rArr` No. of revolution in ONE second `v=(v_(n))/(2pi r_(n))=(v_(2))/(2pi r_(2))` `:.v=(1.095xx10^(6))/(2xx3.14xx2.12xx10^(-10))` `0.0822xx10^(16)s^(-1)` `rArr` No. of revolution in `10^(-8)` second `N=vxx t` `=0.0822xx10^(16)xx10^(-8)` `:.N=8.22xx10^(6)` revolution. |
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