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Using Bohr's atomic model, derive an equation of the radius of the nth orbit of an electron. |
Answer» Solution :The atomic model of Bohr is shown in the figure.  Let mass of an electron m, charge e, LINEAR speed in `n^(th)` ORBIT` v_(n)` and orbit with radius `r_(n)` and charge on nucleus Ze, where Ze = atomic number of an element. The necessary centripetal force is provided Colombian ATTRACTIVE force between electron and nucleus. Thus. `:. (mv_(n)^(2))/(r_(n))=(1)/(4pi epsi_(0))*((Ze)(e))/(r_(n)^(2))=(ze^(2))/(4pi epsi_(0)r_(n)^(2))` where `epsi_(0)`= permittivity of medium From Bohr hypothesis 2, if the anguli momentum of electron in `n^(th)` orbit is `L_(nm)` `L_(n)=(nh)` `:. mv_(n)r_(n)=(nh)/(pi) ""...(2)` From equation (1), `v_(n)(mv_(n)r_(n))=(Ze^(2))/(4pi epsi_(0))` Putting the vlaue of equation (2), `v_(n)-(nh)(2pi)-(Ze^(2))/(4pi epsi_(0))` [ From equation (2)] `:.v_(n)=(Ze^(2))/(4pi epsi_(0))*(1)/((nh)/(2pi))` `:.v_(n)=(1)/(n)*(Ze^(2))/(2h epsi_(0)) ""...(3)` `:.v_(n) prop(Z)/(n)` where `e, 2h,epsi_(0)` constant Hence orbital velocity in `n^(th)` orbit velocity decreases by a factor of n. For hydrogen atom Z = 1, `v_(n) prop (1)/(n)` Now from equation (2). `r_(n)=(nh)/(2pi mv_(n))` Putting the value of eqution (3). `r_(n)=(nhxxnxx2h epsi_(0))/(2pi mxxZe^(2))` `:. r_(n)=(n^(2)h^(2) epsi_(0))/(pi mZe^(2))` `:.r_(n) prop (n^(2))/(Z)[ :.` all other terms are contant] For hydrogen Z=1 `:. r_(n)=(n^(2)h^(2) epsi_(0))/(pi me^(2))""...(4)` `:. r_(n) prop n^(2)` [ `:.`All other terms are constant] The SIZE of the INNERMOST orbit (n = 1) is called Bohr radius, represented by the symbol `a_(0)` thus Putting n = 1 in equation (4), `r_(1)=a_(0)=n^(2)((h^(2) epsi_(0))/(pi me^(2)))` putting the values of `hepsi_(0),pi` and. `a_(0)=5.29xx10^(-11)m` or approximately `0.53Å` |
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