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What is the energy, momentum and wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n=2 to n=1? Given that ionization potential is 13.6 eV |
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Answer» Correct Answer - `16.32xx10^(19)J,5.44xx10^(-27)kg m//sec,` 1218Ã… `E_(1)=-13.6eV` `E_(2)=(-13.6)/(4)eV` `DeltaE=(3)/(4)xx13.6eV` `=0.75xx13.6xx1.6xx10^(-19)J=1.632xx10^(-18)J` `(hc)/(lamda)=1.632xx10^(-18)` `lamda=(6.626xx10^(-34)xx3xx10^(8))/(1.632xx10^(-18))=1218xx10^(-10)m=1218`Ã… `lamda=(h)/(p)` `thereforep=(h)/(lamda)=(6.626xx10^(-34))/(1218xx10^(-10))=5.44xx10^(-27)kg-m//sec`. |
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