Question

Demonstrate that the frequency `omega` of a photon emerging when an electron jumps between neighbouring circular orbits of a hydeogen-like ion satisfies the inequality `omega_(n)gt omegagt omega_(n+1)`, where `omega_(n)` and `omega_(n+1)` are the frequencies of revolution of that electron around the nucleus along the circular orbits. Make sure that as `n rarr oo` the frequency of the proton `omega rarr omega_(n)`

Answer 1

In a circular orbit we have the following formula
`(mv^(2))/(r )=(Ze^(2))/((4pi epsilon_(0))r^(2))`
`mvr= n ħ`
Then `v=(Ze^(2))/((4 pi epsilon_(0))n ħ)`
`r=(n^(2))/(Z)( ħ(4pi epsilon_(0)))/(me^(2))`
The energy `E` is
`E_(n)=(1)/(2)mv^(2)-(Ze^(2))/((4pi epsilon_(0))r`
`=((Ze^(2))/(4pi epsilon_(0)))^(2)(m)/(2 ħ^(2)n^(2))-((Ze^(2))/(4pi epsilon_(0)))^(2)(m)/ ( ħ^(2)n^(2))=m((Ze^(2))/(4pi epsilon_(0)))^(2)//2 ħ^(2)n^(2)`
and the circular frequency of this orbit is
`omega_(n)=(v)/(r )=((Ze^(2))/(4pi epsilon_(0))(m)/(2 ħ^(2)))((1)/(n^(2))-(1)/((n+1)^(2)))`
Thus the inequality
`omega_(n)gtomegagtomega_(n+1)`
will result if
`(1)/(n^(3))gt(1)/(2)((1)/(n^(2))-(1)/((n+1)^(2)))gt(1)/((n+1)^(3))`
Or multiplying by `n^(2)(n+1)^(2)` we have to prove
`((n+1)^(2))/(n)gt(1)/(2)(2n+1)gt(n^(2))/(n+1)`
This can be written as
`n+2+(1)/(n)gtn+(1)/(2)gtn+1-2+(1)/(n+1)`
This is obvious beacuse `-1+(1)/(n)larr(1)/(2)` since `nge1`
For large `n`
`(omega_(n))/(omega_(n)+_1)=((n+1)/(n))^(3)=1+(3)/(n)`
So `(omega_(n))/(omega_(n)+1)rarr1` and we may say `(omega)/(omega_(n))rarr1`