Radiation of atomic hydrogen falls normally on a diffraction grating width `l= 6.6 mm`. The `50th` line of the Balmer series in the observed specturem is close to resolution at a diffraction angle `theta` (in accordace with Rayleigh's criterion). Find the angle.

Explanation

For Balmer series
` ħ omega_(n)= ħR((1)/(4)-(2)/(n^(2))),n ge 3`,
where ` ħR=E_(H)= 13.65eV`. Thus
`(2pi ħc)/(lambda_(n))= ħR((1)/(4)-(1)/(n^(2)))`
or `(2pi ħc)/(lambda_(n+1))-(2pi ħc)/(lambda_(n))= ħR((1)/(n^(2))-(1)/((n+1)^(2)))`
`=ħR((2n+1)/(n^(2)(n+1)^(2)))~~(2R)/(n^(3)) for ngt gt1`
Thus `(2pi ħc)/(lambda_(n)^(2))deltalambda~~(2R ħ)/(n^(3))`
or `(lambda_(n))/(deltalambda)~~(pi ħcn^(3))/(lambda_(n)R ħ)=(picn^(3))/(lambda_(n)R)`
On the other hand for just resolution in a diffracrtion grating
`(lambda)/(deltalambda)=kk(l)/(d)=(l)/(lambdad)k lambda=(1)/(lambdad)d sin theta=(l)/(lambda)sin theta`
Hence
Substitution gives `theta~~ 59.4^(@)`