A certain species of ionized atoms produces emission line spectral accorduing to the Bohr's model. A group of lines in the spectrum is forming a series in which in the shortest wavelength is `22.79 nm` and the longest wavelength is `41.02 nm`. The atomic number of atom is `Z`.
Based on above information, answer the following question:
The series belongs to

Options

A. 2
B. 3
C. 4
D. 5

Answer

C

Explanation

Let n be the lowest energy level of the given series of lines, then maximum wavelength is given by
`(hc)/(lambda_(max)) = 13.6 xx Z^(2) [(1)/(n^(2)) - (1)/((n + 1)^(2))]`
`(hc)/(lambda_(min)) = 136 xx Z^(2) [(1)/(n^(2)) - (1)/(oo^(2))]`
where `lambda_(max) = 41.02 nm and lambda_(min) = 22.79nm`.
Solving above equations, we get `n = 2`, which corresponds to Balmer series.
`Z = 4`
For next to longest wavelength, transition takes place from `n_(1) = 4 to n_(2) = 2`.
So, `(hc)/(lambda) = 13.6 xx 4^(2) [(1)/(4) - (1)/(16)]`
`:. lambda = 30.47 nm`