`DeltaE=(2pi^(2)me^(4))/(h^(2))[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
प्रश्नानुसार `n_(1)=2` और `n_(2)=5`
`therefore" "DeltaE=2179xx10^(18)[(1)/(2^(2))-(1)/(5^(2))]=4.576xx10^(19)J`
अब, विकिरण की कम्पनावृत, `v=(DeltaE)/(h)`
`=(4.576xx10^(-19)J)/(6.62xx10^(-34)Js)`
`=6912xx10^(15)s^(-1)`
`=6.912xx10^(14)s^(-1)`
विकिरण का तरंग-दैधर्य, `lamda=c/v=(3xx10^(8)ms^(-1))/(6.912xx10^(14)s^(-1))`
`0.434xx10^(-6)m=4.34xx10^(-7)m.`
