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

  • 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 :

    \[X(g)\xrightarrow[{}]{{}}{{X}^{+}}(g)+{{e}^{-}}\] 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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