Using Bohr’s equation for the energy levels of the electron in Hydrogen atom, determine the energy of an electron in n = 4

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$- 5.45 \times 10^{- 19} J$

$- 1.365 \times 10^{- 19} J$

$1.84 \times 10^{- 29} J$

$- 1.84 \times 10^{- 29} J$

answer is C.

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Detailed Solution

$E_{n} = \frac{- 13.6}{n^{2}} ev 1 ev = 1.6 \times 10^{- 19 J}, n = 4 E_{n} = \frac{- 13.6}{{(4)}^{2}} \times 1.6 \times 10^{- 19} E_{n} = - 1.365 \times 10^{- 19} J$

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