Finf the wavelength of the first line of the `HE^(+)`ion spectral series whose interval between the extreme line is `Delta omega= 5.18.10^(15) s^(-1)`.

Explanation

We start from the generalized Balmer formula
`omega=RZ^(2)((1)/n^(2)-(1)/(m^(2)))`
Here `m=n+1, n+2,....oo`
Then the angular frequency of the first line of this series (series `n`) is
`omega_(1)=RZ^(2)((1)/(n^(2))-(1)/((n+1)^(2)))=Delta omega(((n+1)/(n))^(2)-1)`
`=Delta omega [{(Zsqrt((R)/(Delta omega)))/(Zsqrt((R)/(Deltaomega)))}-1]`
Then the wavelength will be
`lambda_(1)=(2pi c)/(omega_(1))=(2pi c)/(Delta omega)((Zsqrt((R)/(Delta omega)-1)))/(2Zsqrt((R)/(Delta omega))-1)`
Substitution (with the value `R` from problem 6.34 which is also the correct value determined directly) gives
`lambda_(1)= 0.468mu m`.