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  5. Classification of Elements and Periodicity in Properties / तत्त्वों का वर्गीकरण एवं गुणों में आवर्ति

JEE Main & Advanced Chemistry Classification of Elements and Periodicity in Properties / तत्त्वों का वर्गीकरण एवं गुणों में आवर्ति Sample Paper Topic Test - Periodic Table

  • question_answer
    \[{{N}_{0}}/2\] atoms of X(g) are converted into \[{{X}^{+}}(g)\] by energy E. \[{{N}_{0}}/2\] 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) \[\frac{({{E}_{1}}-{{E}_{2}})}{{{N}_{0}}},\frac{2{{E}_{2}}}{{{N}_{0}}}\]

    D) None is correct

    Correct Answer: B

    Solution :

    [b] \[X(g)\xrightarrow[{}]{{}}{{X}^{+}}(g)+{{e}^{-}}\]
    If I is ionisation energy, then
    \[\therefore \]            \[\frac{{{N}_{0}}}{2}(I)={{E}_{1}}\]            \[\therefore \]             \[I=\frac{2{{E}_{1}}}{{{N}_{0}}}\]
    If E is electron affinity, then
                \[X(g)+{{e}^{-}}\xrightarrow[{}]{{}}{{X}^{-}}(g)\]
    \[\frac{{{N}_{0}}}{2}(E)={{E}_{2}}\]
    \[\therefore \]            \[E=\frac{2{{E}_{2}}}{{{N}_{0}}}\]


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