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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 ? |
Answer» 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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