Calculate the shortest wavelength of Balmer series in Hydrogen atom spectrum. Given `R= 10970000 m^(-1)` .

Explanation

The wavelength of different member of Balmer series is given by
Wavelength will be minimum when `n_(i) =infty`
`(1)/(lambda_(min))=(R_(H))/(4)`
we know Rydberg constant
`=R_(H)=1.097 xx10^(7)`
We know Rydberg constant
`(1)/(lambda_(min))=(4)/(1.097xx10^(7))`
`lambda_(mini)=3464 A`
`therefore` The shortest wavelength of balmer series =3646 A