EAMCET Medical EAMCET Medical Solved Paper-2005

  • question_answer
    According to Moseleys law, the frequency v of the K\[\alpha \] line and the atomic number Z of the element have the relation:

    A)  \[v\,\alpha \left( Z-b \right)\]                                   

    B)  \[v\,\alpha {{\left( Z-b \right)}^{2}}\]

    C)   \[\sqrt{v}\,\alpha {{\left( Z-b \right)}^{2}}\]                     

    D)  \[v\,\alpha \frac{1}{{{\left( Z-b \right)}^{2}}}\]

    Correct Answer: C

    Solution :

                     According to Moseley the frequency of emitted radiation corresponding to the transition of electron from energy level \[{{n}_{2}}\] to \[{{n}_{1}}\]is \[v=R{{(Z-b)}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[\Rightarrow \]               \[v\propto {{(Z-b)}^{2}}\] or            \[\sqrt{v}\propto (Z-b)\]


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