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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

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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