1.

The wave number of the shortest wavelength transition in the Balmer series of H atom is

Answer»

`27419.5cm^(-1)`
`219356cm^(-1)`
`12186.2cm^(-1)`
`24372.4cm^(-1)`

SOLUTION :`barv=109678 [1/(n_(1)^(2))-1/(n_(2)^(2))]`
`n_(1)=2,n_(2)=OO`
`barv=109678 [1/(2^(2))-1/(oo^(2))]`
`=109678/4=27419.5cm^(-1)`


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