How much energy is required to excite an electron from $n_{1} = 1$ to $n_{2} = 2$ in a hydrogen atom? [Given : IE of hydrogen atom is $1.312 \times 10^{6} J (m o l)]$

3.28 \times 10^{5} J m o l^{- 1}

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9.84 \times 10^{5} J m o l^{- 1}

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7.56 \times 10^{5} J m o l^{- 1}

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5.56 \times 10^{5} J m o l^{- 1}

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Solution

The correct option is B $9.84 \times 10^{5} J m o l^{- 1}$
$E_{1} = - 1.312 \times 10^{6} J m o l^{- 1}$
$E_{2} = \frac{- 1.312 \times 10^{6} (1)^{2}}{(2)^{2}} = - 3.28 \times 10^{5} J m o l^{- 1}$
$Δ E = E_{2} - E_{1} = 9.84 \times 10^{5} J m o l^{- 1}$

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Similar questions

1.312 \times 10^{6} J m o l^{- 1}

1.312 \times 10^{6}

m o l^{-}

1.312 \times 10^{6}

J m o l^{- 1}

n_{1} = 1

n_{2} = 2

1.312 \times 10^{6} J {m o l}^{- 1}

1.41 \times 10^{5} J m o l^{- 1}

(i n k J)

O^{-}

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