1.

What is the frequency in Hz of emitted radiation when an electron jumps from fourth orbit to secondth orbit in a hydrogen atom ? [Where R=10^(5)cm^(-1) ]

Answer»

`(3)/(4)xx10^(15)`
`(3)/(16)xx10^(15)`
`(9)/(16)xx10^(15)`
`(3)/(8)xx10^(15)`

Solution :`(1)/(lamda_(ik))=R((1)/(n_(k)^(2))-(1)/(n_(i)^(2)))`
`=10^(5)((1)/(2^(2))-(1)/(4^(2)))""( :.R=10^(5)cm^(-1))`
`=10^(5)((1)/(4)-(1)/(16))`
`=10^(5)((4-1)/(16))`
`(1)/(lamda_(ik))=10^(5)XX(3)/(16)`
but `c=3xx10^(10)cm//s`
`(c)/(lamda_(ik))=(3xx10^(10)xx10^(5)xx3)/(16)`
`:.f_(ik)=(9)/(16)xx10^(15)Hz`


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