JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    \[\frac{{{N}_{0}}}{2}\] atoms of \[X(g)\] are converted into \[{{X}^{+}}(g)\] by energy \[{{E}_{1}}\cdot \frac{{{N}_{0}}}{2}\] atoms of \[X(g)\] are converted into \[{{X}^{-}}(g)\]by energy \[{{E}_{2}}\]. Hence, ionization potential and electron affinity of \[X(g)\] are

    A)  \[\frac{2{{E}_{1}}}{{{N}_{0}}},\,\,\frac{2({{E}_{1}}-{{E}_{2}})}{{{N}_{0}}}\]   

    B)  \[\frac{2{{E}_{1}}}{{{N}_{0}}},\,\,\frac{2{{E}_{2}}}{{{N}_{0}}}\]

    C)  \[\frac{{{E}_{1}}-{{E}_{2}}}{{{N}_{0}}},\,\,\frac{2{{E}_{1}}}{{{N}_{0}}}\]       

    D)  None of these

    Correct Answer: B

    Solution :

     \[\because \,\,X(g)\,\to \,{{X}^{+}}(g)\,+{{e}^{-}};\,\,\Delta H={{E}_{1}}\] for \[\frac{1}{2}\] mol \[\therefore \]                \[IP=\frac{2{{E}_{1}}}{{{N}_{0}}}\] Again \[\because \]     \[X(g)+{{e}^{-}}\to {{X}^{-}}(g),\] \[\Delta H=-{{E}_{2}}\] for \[\frac{1}{2}\] mole \[\therefore \]    \[EA=\frac{2{{E}_{2}}}{{{N}_{0}}}\]

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