Calculate for a hudrogen atom and a `He^(+)` ion:
(a) the radius of the first Bohr orbit and velocity of an electron moving along it ,
(b) the kineric energy and the binding energy of an electron in the ground state,
(c ) the ironizartion potential, the first excitaion potential and the wavelength of the resonance line (`n' =2 rarr=1)`

Explanation

The basic equations have been derived in the problem (6.2). We rewrite them here and determine the required values.
(a)`r_(1)=( ħ^(2))/(m(Ze^(2)//4piepsilon_(0))),Z=1 for Z=2 fo rHe^(+)`
Thus `r_(1)= 52.8p m`,for H atom
`r_(1)= 26.4p m,fo r He^(+)ion`
`v_(1)=(Ze^(2))/((4pi epsilon_(0)) ħ)`
`v_(1)= 2.191xx19^(6)m//s` for H atom
`=4.382xx10^(6)m//s` fo r `He^(+) ion`
(b) `T= (1)/(2)mv_(1)^(2)=(m(Ze^(2))^(2))/((4pi epsilon_(0))^(2)2 ħ^(2))`
`T= 13.65eV` for H atom
`T= 54.6eV` for `He^(+)`ion
In both case `E_(b)=T` beacuse `E_(b)= -E` and `E= -T` (Recall that for coulomb force `V=-2T`)
(c ) The ironization potential `varphi_(i)` is given by
`e varphi_(i)=E_(b)`
so `varphi_(i)= 13.65vol ts` for H atoms
`varphi_(i)= 54.6` volts for `He^(+)`inon
The energy level are `E_(n)=(13.65)/(n^(2))eV` for H atoms
and `E_(n)=-(54.6)/(n^(2))eV` for `He^(+)ion`
Thus `varphi_(1)=13.65(1-(1)/(4))` Volts `=10.23` volts for H atom
`varphi_(1)= 4xx10.23= 40.9` volts for `H^(+) ion`
The wavelength of the resonance line
`(n'=2rarrn=1)` is given by
`(2pi ħc)/(lambda)=-(13.6)/(4)+(13.6)/(1)= 10.23eV` for H atom
so `lambda= 121.2 nm` for H atoms
For `He^(+) ion lambda=(121.2)/(4)=30.3nm`