The energy of an electron in the $2^{n d}$ Bohr orbit of H- atom is –E. The energy of the electron in the Bohr’s first orbit is;

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-4 E

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-2E

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Solution

The correct option is B -4 E
$E \propto \frac{I}{n^{2}}, E_{1} \propto \frac{1}{1} a n d E_{2} \propto \frac{1}{4}, \frac{E_{1}}{E_{2}} = \frac{4}{1} o r = - 4 E$

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