JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2010

  • question_answer
        The potential difference between the target and the cathode of an X-ray tube is 50 kV and the current in tube is 20 mA. Only 1% of the total energy supplied is emitted as X-radiation. What is the maximum frequency of the emitted radiation?

    A)  \[1.2\times {{10}^{17}}Hz\]       

    B)  \[1.2\times {{10}^{19}}Hz\]

    C)  \[6\times {{10}^{15}}Hz\]           

    D)  \[2.4\times {{10}^{18}}Hz\]

    Correct Answer: B

    Solution :

                    Maximum frequency of emitted radiation \[{{v}_{\max }}=\frac{eV}{h}\]                 \[=\frac{1.6\times {{10}^{-19}}\times 50\times {{10}^{3}}}{6.6\times {{10}^{-34}}}\]                 \[=1.2\times {{10}^{19}}Hz\]


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