JEE Main & Advanced Sample Paper JEE Main - Mock Test - 22

  • question_answer
    If \[{{K}_{\alpha }}\]-radiation of Mo (Z = 42) has a wavelength \[0.71\overset{\text{o}}{\mathop{\text{A}}}\,\], the wavelength of the corresponding radiation of Cu(Z = 29) is equal to

    A) \[1.52\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B) \[0.71\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) \[0.36\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D) \[2.5\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: A

    Solution :

    [a] : From Moseley?s law, as \[\lambda \propto \frac{1}{{{(Z-1)}^{2}}}\] \[\frac{{{\lambda }_{Mo}}}{{{\lambda }_{MCu}}}=\frac{({{Z}_{cu}}-{{1}^{2}})}{{{({{Z}_{Mo}}-1)}^{2}}}=\frac{{{(29-1)}^{2}}}{{{(42-1)}^{2}}}=0.4663\] or \[{{\lambda }_{Cu}}=\frac{{{\lambda }_{Mo}}}{0.4663}=\frac{0.71\overset{\text{o}}{\mathop{\text{A}}}\,}{0.4663}=1.52\overset{\text{o}}{\mathop{\text{A}}}\,\]


You need to login to perform this action.
You will be redirected in 3 sec spinner