A) \[8.82\times {{10}^{-17}}J\text{ }ato{{m}^{-1}}\]
B) \[4.41\times {{10}^{-16}}J\text{ }ato{{m}^{-1}}\]
C) \[-4.41\times {{10}^{-17}}J\text{ }ato{{m}^{-1}}\]
D) \[-2.2\times {{10}^{-15}}J\text{ }ato{{m}^{-1}}\]
Correct Answer: C
Solution :
\[\frac{I.{{E}_{1}}}{I.{{E}_{2}}}=\frac{Z_{1}^{2}}{Z_{2}^{2}}\] \[=\frac{19.6\times {{10}^{-18}}}{x}=\frac{4}{9}\] \[x=\frac{9}{4}\times 19.6\times {{10}^{-18}}=44.1\times {{10}^{-18}}J/atm.\] \[=4.41\times {{10}^{-17}}J/atm\]
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