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JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    The wavelength of Ka X-ray of certain material is 12.42 pm. It takes 10 keV to knock out the electron from M shell of this A atom. What should be the minimum accelerating potential across an X-ray tube having A target, which allows product of \[{{K}_{\alpha }}X\]-rays?

    A)  100 keV                              

    B)  90 keV

    C)  110 keV                              

    D)  None of these

    Correct Answer: C

    Solution :

    \[{{E}_{K}}-{{E}_{M}}=\frac{1242}{12.42}keV=100keV\]  [Energy for \[{{K}_{\beta }}X-ray\]]  [Energy required to take the \[{{e}^{-1}}\] from M shell to infinity] So,       \[{{E}_{K}}-0=10keV\] [Here, we are considering only the values without sign]


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