Find the Rydberg correction for the 3P term of a Na atom whose first exitation potential is 2.10V and whose valance electron in the normal 3S state has the binding energy 5.14 eV.

Class 12 Physics in Class 12 3 years ago

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Solution :The ENRGY of the `3p` state must be `-(E_(0)-E varphi)` where `-E_(0)` is the ENERGY of the `3S` state. Then
`E_(0)-evarphi_(1)=( ħR)/((3+alpha_(1))^(2))`
so `alpha_(1)=sqrt(( ħR)/(E_(0)-e varphi_(1)))-3 =-0.885`

Posted on 11 Dec 2021, this text provides information on Class 12 related to Physics in Class 12. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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