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Using the formula for the radius of nth orbit r_(n)=(n^(2)h^(2)epsi_(0))/(pi mZe^(2)) derive an expression for the total energy of electron in n^(th) Bohr's orbit. |
Answer» Solution :The atomic model of Bohr is shown in the figure.  Let mass of electron m, charge e, linear speed in `n^(th)` orbit `v_(n)` and orbital radius `r_(n)`. POSITIVE charge on nucleus Ze, where Z = atomic number of element. The necessary centripetal force is provided by Colombian attractive force between an electron and the positive charge of the nucleus. `:.(mv_(n)^(2))/(r_(n))=((Ze)(e))/(4pi epsi_(0)r_(n)^(2))` `:.(1)/(2)mv_(n)^(2)=(Ze^(2))/(8 pi epsi_(0)r_(n))` `:.` Kinetic energy `K=(Ze^(2))/(8pi epsi_(0)r_(n)) ""....(1)` And potential energy. `U=(kq_(1)q_(2))/(r_(n))` `:.U=-((Ze)(e))/(4pi epsi_(0) r_(n))=-(Ze^(2))/(4pi epsi_(0)r_(n))....(2)` `rArr` Total energy of electron, `E_(n)` = kinetic energy K + potential energy `=(Ze^(2))/(8 pi epsi_(0)r_(n))-(Ze^(2))/(4pi epsi_(0)r_(n))`[ `:.` from equation (1) and (2)] `=-(Ze^(2))/(8pi epsi_(0)r_(n))` Now putting `r_(n)=(n^(2)h^(2)epsi_(0))/(pi m Ze^(2))` `E_(n)=-(Ze^(2))/(8pi epsi_(0))XX(pi m Ze^(3))/(n^(2)h^(2)epsi_(0))` `:. E_(n)=(mZ^(2)e^(4))/(8n^(2)h^(2)epsi_(0)^(2)) ""...(3)` `:. E_(n) prop -(Z^(2))/(n^(2))`[ `:.` All other terms are contant] For hydrogen atom Z=1. `E_(n)=(me^(4))/(8 n^(2)h^(2) epsi_(0)^(2)) ""...(4)` which is the total energy of electron in the `n^(th)` orbit of atom. It is assumed that deriving this formula the electronic orbits are circular. Putting the accepted value of m, e, h and €, in equation (4), `E_(n)=(2.18xx10^(-18))/(n^(2))J` but `1.6xx10^(-19)KJ=1eV` `:.E_(n)=-(2.18xx10^(-18))/(n^(2)xx1.6xx10^(-19))eV` `=-(13.6)/(n^(2))eV [ "where" (me^(4))/(8h^(2) epsi_(0)^(2))=3.6eV]` The negative sign of the total energy INDICATES that electron is bound with nucleus. |
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