JEE Main · ChemistryEasy

$The energy of an electron in the first Bohr orbit of the hydrogen atom is E_{1} = - 13.6 eV . What is the energy of an electron in the second Bohr orbit E_{2} ?$

A. $- 3.4 eV$
B. $- 6.8 eV$
C. $- 27.2 eV$
D. $- 1.51 eV$

Show correct answer & step-by-step solution

Correct answer: A — $- 3.4 eV$

Solution

$The energy formula is E_{n} = \frac{- 13.6}{n ^{2}} eV .$
$For the second orbit, n = 2 .$
$E_{2} = \frac{- 13.6}{2 ^{2}} eV .$
$E_{2} = \frac{- 13.6}{4} eV = - 3.4 eV .$
$Hence the answer is (A).$

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