Calculate the Rydberg constant ` R_H` if `He^+` ions are known to have the wavelength difference between the from ( of the longest wavength ) lines fo Balmer and Lyman series equal to ` 133.7 nm`.

Explanation

For `He^(+)` ion spectrum
`1/lambda = RZ^2 (1/(n_1^2) - 1/(n_2^2))`
where R is Rydberg constant and Z = 2 (for `He^+` ion) For longest wavelength line in Balmer series,
`n_1 = 2, n_2 3`
`1/(lambda_B) = R(2)^2 (1/(2^2) - 1/(3^2))`
`= 4R (1/4 - 1/9) = (4R xx 5)/(36)`
or `" " lambda_B = 9/(5R)`
For longest wavelength line in Lyman series,
`n_1 = 1, n_2 = 2`,
`1/(lambda_L) = R(2)^(2) (1/(1^2) - 1/(2^2))`
`= 4R (1/1 - 1/4) = (4R xx 3)/(4)`
or `lambda_L = 1/(3R)`
Now, `lambda_B - lambda_L = 133.8 xx 10^(-9) m`
`133.8 xx 10^(-9) = 9/(5R) - 1/(3R)`
`= 1/R (9/5 - 1/3) = 1/R ((27 - 5)/(15))`
or `133.8 xx 10^(-9) = 1/R xx 22/15`
`R = (22)/(15 xx 133.8 xx 10^(-9)) = 1.0961 xx 10^(7) m^(-1)`.