Question

Given that for `He^(+)` ion, the difference between the longest wavelength line of Balmer series and Lymen series is 133.8 nm, calculate the value of Rydberg constant.

Answer 1

For H-like particle.
`bar(v) = (1)/(lamda) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
where R is Rydberg constant
For `He^(+), Z = 2` and for longest wavelength line of Balmer series, `n_(1) = 2, n_(2) =3`. Hence,
`(1)/(lamda_("Balmer")) = R (2)^(2) ((1)/(2^(2)) - (1)/(3^(2))) = 4R ((1)/(4) - (1)/(9)) = 4R xx (5)/(36) = (5R)/(9) or lamda_("Balmer") = (9)/(5R)`
For longest wavelength in Lyman series, `n_(1) =1, n_(2) = 3=2`. Hence,
`lamda_("Lyman") = R (2)^(2) ((1)/(1^(2)) - (1)/(2^(2))) = 4R (1 - (1)/(4)) = 4R xx (3)/(4) = 3R or lamda_("Lyman") = (1)/(3R)`
Given `lamda_("Balmer") - lamda_("Lyman") = 133.8 xx 10^(-9)m`
`:. (9)/(5R) - (1)/(3R) = 133.8 xx 10^(-9) or (1)/(R) ((9)/(5) - (1)/(3)) = 133.8 xx 10^(-9)`
or `(1)/(R) xx (22)/(15) = 133.8 xx 10^(-9) or R = (22)/(15) xx (1)/(133.8 xx 10^(-9)) = 1.0961 xx 10^(9) m^(-1)`