Calculate the Rydberg constant `R` if `He^(+)` ions are known to have the wavelength difference between the first (of the longest wavelength) lines of the Balmer and Lyman series equal to `Delta lambda = 133.7nm`.

Explanation

For the longest wavelength (first) line of the Balmer series we have on using the generalized Balmer formula
`omega= Z^(2)R((1)/(n^(2))-(1)/(m^(2)))`
the resultant `lambda_(1 Lyman)=(2pic)/(Z^(2)R(1-(1)/(4)))=(8pi c)/(3Z^(2)R)`
Then `Delta lambda=lambda_(1Balmar)-lambda_(1Lymon)=(176 pic)/(15Z^(2)R)`
so, `R=(176 pi c)/(15Z^(2)Delta lambda)= 2.07xx10^(16)sec^(-1)`