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JEE Main & Advanced Sample Paper JEE Main Sample Paper-42

  • question_answer
    \[{{\text{X}}_{\text{ }\!\!\beta\!\!\text{ }}}\]X-ray of argon has wavelength 0.36 nm, the minimum energy needed to ionize an argon atom is 16 eV. Find the energy needed to knock out an electron from K shell of an argon atom.

    A)  16 eV                         

    B)  3.45 keV

    C)  3.47 keV       

    D)  3.43 keV

    Correct Answer: C

    Solution :

    \[{{E}_{K}}-{{E}_{M}}=\frac{1242}{0.36}=3450eV\]  (Energy corresponding to\[{{K}_{\beta }}X-ray\]) \[{{E}_{M}}-0=16eV\]  (lionization energy) from M shell. The energy needed to knock out an \[{{e}^{-1}}\] from K shell is \[{{E}_{K}}=3450+{{E}_{M}}\] \[{{E}_{K}}=3466eV\] \[=3.466keV\approx 3.47keV\]


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