\[{{N}_{0}}/2\] 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

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

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

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

\[{{N}_{0}}/2\] atoms of X(g) are converted into \[{{X}^{+}}(g)\] by e

1.	\[{{N}_{0}}/2\] 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
Answer» C. \[\frac{({{E}_{1}}-{{E}_{2}})}{{{N}_{0}}},\frac{2{{E}_{2}}}{{{N}_{0}}}\]

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\[{{N}_{0}}/2\] 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

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