A) -13.6eV
B) -3.4eV
C) -0.85eV
D) -1.70eV
Correct Answer: D
Solution :
[d] Energy of nth orbit of H-atom \[=-\frac{2{{\pi }^{2}}m{{e}^{4}}{{k}^{2}}}{{{h}^{2}}}\times \frac{1}{{{h}^{2}}}\] Energy of Boht's orbit of H-atom \[=-\frac{2{{\pi }^{2}}m{{e}^{4}}{{k}^{2}}}{{{h}^{2}}}\] \[=-13.6eV\](given) Energy of fourth Bohr's of H-atom \[=-\frac{2{{\pi }^{2}}m{{e}^{4}}{{k}^{2}}}{{{h}^{2}}}\times \frac{1}{{{4}^{2}}}\] \[=13.6\times \frac{1}{16}eV=-0.85eV\] PE of electron in nth orbit =\[2\times {{E}_{n}}\] So P.E. of electron in \[{{4}^{th}}\]orbit \[=2\times (-0.85)=-1.70eV\]
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