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On the basis of Bohr's theory, calculate the velocity and time period of revolution of the electron in the innermost orbit (n = 1) of the hydrogen atom. Given : Bohr's Radius (r_(1)) = 0.53 Å. Data: n=1, r_(1)=0.53, Å=0.53xx 10^(-10) m, v= ?, T=? |
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Answer» Solution :KINETIC energy of the electron `=(1)/(2)mv_(n)^(2)` `=(Ze^(2))/(8pi epsi_(0)r_(n))` `(1)/(2) mv_(n)^(2)=(Ze^(2))/(8 pi epsi_(0)[(n^(2)h^(2)epsi_(0))/(pi m Ze^(2))]) [ :.r_(n)=(n^(2)h^(2)epsi_(0))/(pi mZe^(2))]` `rArr v_(n)=(Ze^(2))/(2epsi_(0)nh)` For HYDROGEN atom Z = 1 and for n=1, `v_(1)=(e^(2))/(2epsi_(0)h)=((1.6xx10^(-19))^(2))/(2xx8.854xx10^(-12)xx6.656xx10^(-34))` `=2.1818xx10^(-10) ms^(-1)` [Note : Velocity can also be calculated using Bohr.s first postulate] Period = Time taken for one revolution `=T=(2pi r_(1))/(v_(1))` `[ :. "time"=("distance")/("velocity")]` `:.T=(2xx3.142xx0.53xx10^(-10))/(2.1818xx10^(6))` `T=1.5265xx10^(-16)s` |
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