Hydrogen Atom and Hydrogen Molecule
The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by johann Jakob Balmer, a Swiss teacher. Balmer's empirical empirical formula is
`1/lambda=R_(H)(1/2^(2)-1/n^(2))," " N=3, 4, 5`
`R_(H)=(Me e^(4))/(8 epsilon_(0)^(2)h^(3) c)=109. 678 cm^(-1)`
Here,
`R_(H)` is the Rydberg Constant, `m_(e)` is the mass of electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom//ion.
Calculate the iongest wavelength in `A (1 Å=10^(-10) m)` in the 'Balmer series of singly ionized helium `(He^(+))` Ignore nuclear motion in your calculation.

Answer

Longest wavelength `A_(L)` corresponds to `n=3`
For `He^(+)`
`1/lambda=4R_(H)(1/2^(2)-1/n^(2))`
`lambda_(L)=1641.1 Å`