Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n,show that this frequency equals the lassical frequency of revolution of the electron in the orbit.

Explanation

According to hypothesis-2 of Bohr an angular momentum of electron is quantized that is
`mvr=(nh)/(2pi)` [For h aotm Z=1]
`:. nr^(2)omega=(nh)/(2pi)[ :. V=r omega]`
`:.omega=(nh)/(2pi mr^(2))...(1)`
but `r=(n^(2)h^(2)epsi_(0))/(pi me^(2))`
`:. omega=(pi me^(4))/(2n^(3)h^(2)epsi_(0)^(2))`
`:. 2pi=(pi me^(4))/(2n^(3)h^(3)epsi_(0)^(2))`
`:.v=(me^(4))/(4pi n^(3)h^(3)epsi_(0)^(3))`
`:.v=(me^(4))/(4n^(3)h^(3)epsi_(0)^(2)`
Now `K=(1)/(4pi epsi_(0))`
`:. epsi_(0)^(2)=(1)/(16pi^(2)k^(2))`
`:.v=(me^(4))/(4n^(3)h^(3))xx(16pi^(2)k^(2))/(1)`
`:.v=(4pi^(2)k^(2)me^(4))/(n^(3)h^(3)) " "...(2)`
`rArr` When an electron transist from n + 1 ort to `n^(th)` orbit, the formula for wavelength emitted radiation is
`(1)/(lambda)=R[(1)/(n^(2)-(1)/((n+1)^(2)))]`
`=R[(1)/(n^(2))-(1)/(n^(2)(1+(1)/(m))^(2))]`
`=R[(1)/(mn^(2))-(1)/(n^(2))xx(1+(1)/(n))^(-2)]`
But n is greater, so `(1)/(n) lt lt 1`, so keeping first two terms of binomial expansion
`(1)/(lambda)=R[(1)/(n^(2)-(1)/(n^(2))(1-(2)/(n)))]`
`=R[(1)/(n^(2))-(1)/(n^(2))+(2)/(n^(3))]`
`=R[(1)/(n^(2))-(1)/(n^(2))+(2)/(n^(3))]`
`(1)/(lambda)=(2R)/(n^(3))`
but Rydberg constant,
`R=(me^(4))/(8 epsi_(0)^(2)ch^(3))`
`:.(1)/(lambda)=(2me^(4))/(8 epsi_(0)^(2)ch^(3)n^(3))=(me^(4))/(4 epsi_(0)^(2)ch^(3)n^(2))`
`:.(c)/(lambda)=(me^(4))/(4epsi_(0)^(2)h^(3)n^(3))`
putting in `epis_(0)^(2)=(1)/(16pi^(2)k^(2))`
`v=(4pi^(2)k^(2)me^(4))/(n^(3)h^(3)) " "...(3)`
Both equation (2) and (3) are equal.
`:.` Hence, emitted frequency according Bohr theory is equal to its classical orbital frequency due to the transition of electron.