A) \[-2.18\times {{10}^{-18}}\text{ }J\text{ }ato{{m}^{-1}}\]
B) \[-2.18\times {{10}^{-18}}\text{ k}J\text{ }ato{{m}^{-1}}\]
C) \[-2.18\times {{10}^{-18}}\text{ ergs }ato{{m}^{-1}}\]
D) \[+2.18\times {{10}^{-18}}\text{ J }ato{{m}^{-1}}\]
Correct Answer: A
Solution :
\[{{E}_{n}}=-\frac{13.6\,{{Z}^{2}}}{{{n}^{2}}}eV\,\,\,ato{{m}^{-1}}\] \[{{E}_{n}}=-\frac{2.18\times {{10}^{-18}}{{Z}^{2}}}{{{n}^{2}}}J\,\,ato{{m}^{-1}}\] For first orbit of hydrogen atom n = 1 and Z = 1 \[{{E}_{1}}=-\frac{2.18\times {{10}^{-18}}{{(1)}^{2}}}{{{(1)}^{2}}}J\,\,ato{{m}^{-1}}\] \[{{E}_{1}}=-2.18\times {{10}^{-18}}\,J\,\,ato{{m}^{-1}}\]
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