A gas of hydrogen like ions is prepared in such a way that ions are only in the ground state and the first excited state. A monochromatic light of wavelength 1216 Å is absorbed by the ions. The ions are lifted to higher excited states and emit radiations of six wavelengths. Find the principal quantum number of the final excited state.

Explanation

The energy state of hydrogen-like atom is given by
`E_(n) = - (B)/(n^(2))`
where `B = Z^(2) Rhc`.
The emission of six wavelength implies that the higher excited state corresponds to quantum number `n = 4` . As some wavelength are lower and some wavelength are higher , it implies that the initial excited state is `n = 2` . Thus the principal quantum number of excited states are `n = , n = 3, n = 4`.

`:.(hc)/(lambda) = B ((1)/(2^(2)) - (1)/(4^(2))) implies (hc)/(lambda) = (3 B)/(16)`
`:. B = (1.6 xx 6.62 xx 10^(-34) xx 3 xx 10^(8))/(3 xx (1216 xx 10^(-10))`
`= 8.71 xx 10^(-18)J`
If `Z` is atimic number of necleus, then `B = Z^(2) Rhc`
`:. Z^(2) = (B)/(Rhc) = (8.17 xx 10^(-18)J)/(13.6 xx 1.6 xx 10^(-19) = 4`
`:. Z = 2`
The maximum wavelength of emitted redition corresponds to transition `n = 4 to n = 3` and is given by
`(hc)/(lambda _(max)) = B ((1)/(3^(2)) - (1)/(4^(2)))`
`implies lambda _(max) = (144 hc)/(7 B) = (144 xx 6.62 xx 10^(-34) xx 3 xx 10^(8))/(7 xx 8.71 xx 10^(-18))`
`= 4.69 xx 10^(-7) m`
The minimum wavelength of emitted rediction corresponds to transition `n = 4 to n = 1` and is given by
`(hc)/(lambda _(max)) = B ((1)/(1^(2)) - (1)/(4^(2)))`
`implies lambda _(max) = (16 hc)/(15 B) = (16 xx 6.62 xx 10^(-34) xx 3 xx 10^(8))/(15 xx 8.71 xx 10^(-18))`
`= 0.2 xx 10^(-7) m`