JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Critical Thinking

  • question_answer
    The X-ray wavelength of \[{{L}_{\alpha }}\] line of platinum (Z=78) is \[1.30{AA}.\] The X ?ray wavelength of \[{{L}_{\alpha }}\] line of Molybdenum (Z=42) is                                                [EAMCET (Eng.) 2000]

    A)  5.41 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B) 4.20 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) 2.70 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]                                      

    D) 1.35 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: A

    Solution :

    The wave length of \[{{L}_{\alpha }}\]line is given by                    \[\frac{1}{\lambda }=R\,{{(z-7.4)}^{2}}\,\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)\,\Rightarrow \lambda \propto \frac{1}{{{(Z-7.4)}^{2}}}\]                    \[\Rightarrow \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{{{({{z}_{2}}-7.4)}^{2}}}{{{({{z}_{1}}-7.4)}^{2}}}\]\[\Rightarrow \frac{1.30}{{{\lambda }_{2}}}=\frac{{{(42-7.4)}^{2}}}{{{(78-7.4)}^{2}}}\]\[\Rightarrow \,{{\lambda }_{2}}=5.41\,{\AA}\]


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