A) 8.53
B) 7.51
C) \[8.92\]
D) 9.86
Correct Answer: D
Solution :
\[{{E}_{2}}=\frac{-1.312\times {{10}^{6}}\times {{(1)}^{2}}}{{{(2)}^{2}}}\] \[\text{=-3}\text{.28 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{J/mol}\] \[{{\text{E}}_{\text{1}}}\text{=-1}\text{.312 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\text{J/mol}\] \[\therefore \] \[\Delta E={{E}_{2}}-{{E}_{1}}\] \[=-3.28\times {{10}^{6}}-(1.312\times {{10}^{6}})\] \[\text{= 9}\text{.84 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{J/mol}\]
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