The ionisation energy of He- is 19-6- 10-18 J atom -1- The energy of the stationary state of Li2- will be-

The correct option is B 44.10× 10^ 18 J /atom ...

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The ionisation energy of $H e^{+}$ is $19.6 \times 10^{- 18}$ J atom $^{- 1}$ . The energy of the stationary state of $L i^{2 +}$ will be:

84.2 \times 10^{- 18}

J /atom

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44.10 \times 10^{- 18}

J /atom

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63.2 \times 10^{- 18}

J /atom

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21.2 \times 10^{- 18}

J /atom

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Solution

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Ionisation energy of He⁺ is 19.6×10^-18 J atom^-1. The energy of the first stationary state(n=1) of Li²⁺ is:

19.6 \times 10^{- 18} J a t o m^{- 1}

L i^{2 +}

H e^{+}

8.71 \times 10^{- 18} J a t o m^{- 1}

L i^{2 +}

{H e}^{+}

19.6 \times 10^{- 18} J p e r a t o m

{L i}^{2 +}

H e^{+}

19.6 \times 10^{- 18}

L i^{2 +}

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