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The ionization energy of hydrogen atom is $- 13.6 eV .$ The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is (Avogadro's constant $= 6.022 \times 10^{23})$

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$1.69 \times 10^{25} J$

$1.69 \times 10^{- 20} J$

$1.69 \times 10^{- 23} J$

$1.69 \times 10^{23} J$

answer is B.

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Detailed Solution

$E = \frac{- 13.6}{n^{2}} = \frac{- 13.6}{4} = - 3.4 eV$

We know that energy required for excitation

$Δ E = E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV .$

Therefore energy required for excitation of electron per

atom $= \frac{10.2}{6.02 \times 10^{23}} = 1.69 \times 10^{- 23} J .$

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