JEE Main & Advanced Sample Paper JEE Main - Mock Test - 11

  • question_answer
    The wavelength \[{{K}_{\alpha }}\]of X-rays for two metals 'A and 'B' are \[\frac{4}{1875\,R}\] and \[\frac{1}{675\,R}\]respectively, where 'R' is Rydberg constant. Find the number of elements lying between A and B according to their atomic numbers

    A) 3     

    B) 1       

    C) 4                     

    D) 5

    Correct Answer: C

    Solution :

    Using \[\frac{1}{\lambda }=R{{(Z-1)}^{2}}\left[ \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right]\]
    For a particle, \[{{n}_{1}}=2,\,{{n}_{2}}=1\]
    Form metal A: \[\frac{1875R}{4}=R{{({{Z}_{1}}-1)}^{2}}\left( \frac{3}{4} \right)\Rightarrow {{Z}_{1}}=26\]
    Form metal B: \[675\,R=R{{({{Z}_{2}}-1)}^{2}}\left( \frac{3}{4} \right)\Rightarrow {{Z}_{2}}=31\]
    Therefore, 4 elements lie between A and B
    \[=220\,MeV.\]


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