1.

The wavelength of the radiation emitted when the electron jumps from 4th shell to 2nd shell is

Answer»

`4862 Å`
`2056Å`
`5241 Å`
`109700` CM

Solution :ACCORDING to Balmer EQUATION
Wave number `(barv)=109677(1/n_(1)^(2)-1/n_(2)^(2))cm^(-1)`
`barv=109677(1/((2)^(2))-1/((4)^(2)))cm^(-1)`
`=(109677xx3)/16cm^(-1)`
`lambda=1/v=16/(109677xx3)=4862xx10^(-8)`cm
`=4862xx10^(-10)m=4862 Å`


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