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The energy difference between ground state of an atom and its excited state is `3xx10^(-19)J`. What is the wavelength of photom required for this radi

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The energy difference between ground state of an atom and its excited state is `3xx10^(-19)J`. What is the wavelength of photom required for this radiation ?
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Correct Answer - `6.6xx10^(-7)m`
`DeltaE = (hc)/(lambda)`
`lambda=(hc)/(DeltaE)=((6.626xx10^(-34)Js)(3xx10^(8)ms^(-1)))/((3xx10^(19)J))=6.6xx10^(-7) m`
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