What electronic transition in the `He^(+)` ion would emit the radiation of the same wavelength as that of the first Lyman transition of hydrogen (i.e., for an electron jumping from n=2 to n-1) ? Neglect the reduced- mass effect. Also, calculate second ionisation potential of He and first Bohr orbit for `He^(+)` (`e= 1.6 xx 10^(-19)` coulomb, `m= 9.1 xx 10^(-31)kg, h= 6.626 xx 10^(-34)J.s, c= 2.997 xx 10^(8)` metre/s and `in_(0)= 8.854 xx 10^(-12)"coulomb"^(2)`/newton. `"metre"^(2)`)

Explanation

As `He^(+)` has only one electron, Bohr equations may be applied and the Rydberg constant, R, for H and `He^(+)` will be the same as the reduced -mass effect in neglected. For first Lyman transition for hydrogen (i.e., when an electron jimps from n=2 to n=1), we have
`(1)/(lamda)= RZ^(2) ((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))= R ((1)/(1^(2))-(1)/(2^(2)))=(3R)/(4)`
Since wavelength of the radiation emitted in electronic transition in `He^(+)` is the same as for hydrogen, therefore for `He^(+)` (Z=2), we have, `(1)/(lamda)= RZ^(2) ((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))= 4R ((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))= (3R)/(4)`
`therefore n_(1) and n_(2)` have to be 2 and 4 respectively as
`(1)/(lamda)= 4R ((1)/(2^(2))-(1)/(4^(2)))= (3R)/(4)`
As the second IP of He is the energy required to remove the second electron from first orbit to infinite orbit,
`therefore` second IP of He `= (Z^(2) e^(4)m)/(8 epsi_(0)^(2) h^(2)) ((1)/(1^(2))-(1)/(oo))`
`IP = (2^(2) (1.6 xx 10^(-19))^(4) (9.1 xx 10^(-31)))/(8(8.854 xx 10^(-12))^(2) (6.626 xx 10^(-34))^(2))`
`=8.67 xx 10^(-18)` Joules
Now to calculate `r_(1)` for He (Z=2), we have,
`r_(1)= (epsi_(0)n^(2)h^(2))/(pi mZe^(2))`
`=((8.854 xx 10^(-12)) (1)^(2) (6..626 xx 10^(-34))^(2))/((3.14 ) (9.1 xx 10^(-31))^(2) (1.6 xx 10^(-19))^(2))`
`=2.64 xx 10^(-11)` metres