Ionisation energy of He- is 8-71- 10-18 J atom-1- The energy of the first stationary state n - 1 of Li2- is-

Given, Ionisation energy of He^+ = nbsp;19.6× 10^ 18 Jatom^ 1 We know, E_n∝Z^2/n^2 Here, for both the cases n = 1. Hence, E_n∝ Z^2 ∴E_He^+/E_Li^2+=2^2/3^2 ...

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Standard XI

Chemistry

Bohr's Model of a Hydrogen Atom

Ionisation en...

Question

Ionisation energy of $H e^{+}$ is $8.71 \times 10^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state (n = 1) of $L i^{2 +}$ is:

- 4.41 \times 10^{- 16} J a t o m^{- 1}

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- 1.96 \times 10^{- 17} J a t o m^{- 1}

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- 2.76 \times 10^{- 15} J a t o m^{- 1}

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6.72 \times 10^{- 17} J a t o m^{- 1}

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Solution

The correct option is B $- 1.96 \times 10^{- 17} J a t o m^{- 1}$
Given,
$Ionisation energy of H e^{+} = 19.6 \times 10^{- 18} J a t o m^{- 1}$

We know,
$E_{n} \propto \frac{Z^{2}}{n^{2}}$
Here, for both the cases n = 1.
Hence, $E_{n} \propto Z^{2}$
$∴ \frac{E_{H e^{+}}}{E_{L i^{2 +}}} = \frac{2^{2}}{3^{2}}$
$\Rightarrow \frac{- 8.71 \times 10^{- 18}}{E_{L i^{2 +}}} = \frac{4}{9}$
$\Rightarrow E_{L i^{2 +}} = - \frac{9}{4} [8.71 \times 10^{- 18}]$
$= - 1.96 \times 10^{- 17} J / a t o m$

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Bohr's Model of a Hydrogen Atom

Standard XI Chemistry

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Ionisation energy of He+ is 8.71×10−18 J atom−1. The energy of the first stationary state (n = 1) of Li2+ is:

The correct option is B −1.96×10−17 J atom−1Given, Ionisation energy of He+=19.6×10−18 Jatom−1 We know, En∝Z2n2 Here, for both the cases n = 1. Hence, En∝Z2 ∴EHe+ELi2+=2232 ⇒−8.71 × 10−18ELi2+=49 ⇒ELi2+=−94 [8.71×10−18] =−1.96×10−17 J/atom

Ionisation energy of $H e^{+}$ is $8.71 \times 10^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state (n = 1) of $L i^{2 +}$ is: