Read the pasage given below as well as the adjoining energy level diagram and then answer the questions given after the passage .
As per Bohr's theory og hydrogen atom the energy of an atom in a state corresponding to principal quantum number n is given as :
`E_(n) = - (13.6)/(n^(2)) eV`
The energy of an atom is the least when its electron is revolving in an orbit closets to the nucleus i.e., the one for which n =1. This state is called the ground state. When a hydrogen atom receives energy by processes such as electron collision , the atom may acquire sufficient energy to raise the electron to higher energy states and the atom is then said to be in an excited state . From these excited states the electron can then fall back to a state of lower energy , emitting a photon in the process.
The energy level diagram for the stationary states of a hydrogen atom , compouted from Bohr's relation for energy , is given in Fig.12.02. The principal quantum number n labels the stationary states in the ascending order of energy . Obviously , the highest energy corresponds to `n = oo` and has an energy of 0 eV. This is the energy of the atom when the electron is completely removed `(r= oo)` from the nucleus and is at rest.

Now answer the follwoing questions:
Write the formula for spectral lines the of hydrogen atom in visible light in terms of Rydberg's constant.

Explanation

Wavelenght of spectral lines of Balmer series may be expressed as :
`(1)/(lambda) = bar(v) = R [(1)/((2)^(2)) - (1)/(n^(2))]` , where ` n = 3,4,5`…….
Here, R is Rydberg.s constant having a value `1.097 xx 10^(-7) m^(-1)`.