(a) State the postulates of Bohr's model of hydrogen atom and derive the expression for Bohr radius.
(b) Find the ratio of the longest and the shortest wavelengths amongst the spectral lines of Balmer series in the spectrum of hydrogen atom.

Explanation

Bohr gave following three postulates for hydrogen atom:
1. An electron revolves round the nucleus in certain specifie circular motion in such a stationary orbit is provided by electrostatic force of attraction . Thus,
`(mv_n^2)/(r_n)=1/(4piepsilon_0).e^(2)/r_(n)^2" "...(i)`
2. For an orbit to be stationary (or non - radiating , the angular momentum of the electron must be an integer multiple of `h/(2pi)` , where h is the Planck.s constant. Thus,
`L_(m)=mv_(n)r_(n)=(nh)/(2pi)" "...(ii)`
3. Whenever an electron shifs from one of its specified non - radiating orbit to another such orbit, it emits/absorbs a photon whose energy is equal to the initial and final states. Thus,
`E_(i)-E_f=hv=(hc)/lamda" "...(iii)`
Form (i), we get
`v_(n)^(2)=e^2/(4piepsilon_0mr_n)" ".....(iv)`
and from (ii) , we have
`v_(n)=(nh)/(2pimr_n)" "...(v)`
Squaring (v) and then equating it with (iv) , we get
`(n^2h^2)/(4pi^(2)m^(2)r_n^2)=e^2/(4piepsilon_(0)m.r_n)`
`impliesr_n=(n^2h^2)/(4pi^(2)m^(2))xx(4piin_(0)m)/e^2=(in_0h^2)/(pime^2).n^2`
In stable orbit of hydrogen atom n = 1 and then radius of 1st orbit is called Bohr.s radius `.a_0.` . Obviously `a_(0)=(in_(0)h^2)/(pime^2)` . Substituting values of various constants we find that `a_(0) = 5.29 xx10^(-11)m`
(b) We know wavelength of various spectral lines of Balmer series is given by the relation
`1/lamda=R[1/(2)^(2)-1/((n)^(2))]` where n = 3,4,5,6.....
We obtain the longest wavelength when n = 3 and shortest wavelength when `n = oo`
`implies1/(lamda_("longest"))=R[1/(2)^(2)-1/(oo)^2]=R/4`
`implies(lamda_("longest"))/(lamda_("shortest"))=(R/4)/((5R)/36)=9/5`