A) \[E=\frac{2{{\pi }^{2}}m{{e}^{4}}}{{{h}^{2}}}\frac{1}{{{n}^{2}}}\]
B) \[E=\frac{4{{\pi }^{2}}m{{e}^{2}}}{{{h}^{2}}}\frac{1}{{{n}^{2}}}\]
C) \[E=-\frac{2{{\pi }^{2}}m{{e}^{4}}}{{{h}^{2}}}\frac{1}{{{n}^{2}}}\]
D) \[E=-\frac{4{{\pi }^{2}}m{{e}^{2}}}{{{h}^{2}}}\frac{1}{{{n}^{2}}}\]
Correct Answer: C
Solution :
: The energy of an electron in the\[{{n}^{th}}\]orbit of hydrogen is given by\[{{E}_{n}}=\frac{-2{{\pi }^{2}}m{{e}^{4}}}{{{n}^{2}}{{h}^{2}}}\]
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