When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electromagnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as
`E =- (2pi^(2)m_(e)Z^(2)e^(4))/(n^(2)h^(2))`
Where, `m_(e)=` rest mass of electron
`DeltaE = (E_(n_(2))-E_(n_(1))) = 13.6 xx Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]eV` per atom
This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum
`bar(v) = (1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]m^(-1)`
Where `R_(H) = 1.1 xx 10^(7)m^(-1)` (Rydberg constant)
For Lyman, Balmer, Paschen, Brackett and Pfund series the value of `n_(1) = 1,2,3,4,5` respectively and `n_(2) =oo` for series limit. If an electron jumps from higher orbit n to ground state, then number of spectral line will be `.^(n)C_(2)`. Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass `(mu)`.
`(1)/(mu) = (1)/(m_(N))+ (1)/(m_(e))`
Here, `m_(N)=` mass of nucleus `m_(e)=` mass of electron Answer the following questions
In which of the following region the spectrum of `He^(+)` will be observed in above transition?

Options

A. Ultraviolet
B. Visible
C. Infrared
D. far infrared

Answer

B

Explanation

`n_(1) = 2` i.e. balmer series