The energy required for He-He2-e- is energy required for He-Heg-e- is 10eV atom is-

The correct option is C 13.6eV For monoelectronic species , i.E=13.6×Z^2n^2=13.6×42^2=13.6eV Since , in He^+, electron is present in ...

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Standard XII

Chemistry

Introduction to p Block Elements

The energy re...

Question

The energy required for ${H e}^{+} \to {H e}^{2 +} + e^{-}$ is ______ energy required for $H e \to {H e}_{(g)}^{+} + e^{-}$ is $10 e V$ atom is:

13.6 e V

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54.4 e V

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27.2 e V

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1001.4 e V

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Solution

The correct option is C $13.6 e V$
For monoelectronic species $,$
$i . E = 13.6 \times \frac{Z^{2}}{n^{2}} = \frac{13.6 \times 4}{2^{2}} = 13.6 e V$
Since $,$ in ${H e}^{+},$ electron is present in $2 s$ orbital $.$
$∴$ $(n = 2) .$
$∵$ $10 = 13.6 \times \frac{2^{2}}{n^{2}} \Rightarrow n = 2.$

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Similar questions

Calculate the energy required for the process

The ionization energy for the H atom in the ground state is 2.18 ×10^–18 J atom^–1

Q. If the ionisation energy of hydrogen atom is

13.6 e V

, what will be the ionisation energy of

H e^{+}

and

L i^{2 +}

ions respectively?

Q. Calculate the energy required for the process

{H e}^{+} (g) \to {H e}^{2 +} (g) + e^{-}

.
The ionization energy for the H atom in the ground state is

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

Q. Energy required for the process is :

{H e}^{+} g ⟶ H e^{+ 2} (g) + e^{-}

(The ionization energy for the H-atom in the ground state is

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

)

Calculate the energy required for the process
$H e_{(g)}^{+} \to H e_{(g)}^{2 +} + e^{-}$
The ionization energy for the H atom in the ground state is 2.18 × $10^{- 18} J a t o m^{- 1}$

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The energy required for He+→He2++e− is ______ energy required for He→He+(g)+e− is 10eV atom is:

The correct option is C 13.6eVFor monoelectronic species ,i.E=13.6×Z2n2=13.6×422=13.6eVSince, in He+, electron is present in 2s orbital .∴ (n=2).∵ 10=13.6×22n2⇒n=2.

The energy required for ${H e}^{+} \to {H e}^{2 +} + e^{-}$ is ______ energy required for $H e \to {H e}_{(g)}^{+} + e^{-}$ is $10 e V$ atom is:

The correct option is C $13.6 e V$
For monoelectronic species $,$
$i . E = 13.6 \times \frac{Z^{2}}{n^{2}} = \frac{13.6 \times 4}{2^{2}} = 13.6 e V$
Since $,$ in ${H e}^{+},$ electron is present in $2 s$ orbital $.$
$∴$ $(n = 2) .$
$∵$ $10 = 13.6 \times \frac{2^{2}}{n^{2}} \Rightarrow n = 2.$