Ionization energy of is 19-6 - 10-18 J atom -1- The energy of the first stationary state n -1 of Li 2- isA- 8-82 - 10-17 J atom -1B- 4-41 - 10-16 J atom -1C- -4-41 - 10-17 J atom -1D- -2-2 - 10-15 J atom -1

The correct option is B 4.41 X 10 16 J atom 1I_He^+/I_Li^2+=Z^2_1/Z^2_2 19.6× 10^ 18/I_Li^2+=4/9 I_Li^2+=9/4× 19.6× 10^ 18=44.1× 10^ 18=4.41× 10^ 17 J atom^ 1 ...

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Standard XII

Chemistry

Trend in Ionisation Enthalpy

Ionization en...

Question

Ionization energy of is $19.6 \times 10^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state n = 1 of $L i^{2 +}$ is

–2.2 X 10^-15 J atom^-1

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8.82 X 10^-17 J atom^-1

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4.41 X 10^-16 J atom^-1

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-4.41 X 10^-17 J atom^-1

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Solution

The correct option is C 4.41 X 10^-16 J atom^-1
$\frac{I_{H e^{+}}}{I_{L i^{2 +}}} = \frac{Z_{1}^{2}}{Z_{2}^{2}}$
$\frac{19.6 \times 10^{- 18}}{I_{L i^{2 +}}} = \frac{4}{9}$
$I_{L i^{2 +}} = \frac{9}{4} \times 19.6 \times 10^{- 18} = 44.1 \times 10^{- 18} = 4.41 \times 10^{- 17} J a t o m^{- 1}$

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Similar questions

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

Q. The ionisation energy of

H e^{+}

19.6 \times 10^{- 18}

J atom

^{- 1}

. The energy of the stationary state of

L i^{2 +}

will be:

Q. Ionization energy of

{H e}^{+}

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

. The energy of the first stationary state (

n = 1

) of

{L i}^{2 +}

Q. If the ionisation energy of

{H e}^{+}

19.6 \times 10^{- 18} J

per atom then the energy of

{B e}^{3 +}

in the

2^{n d}

stationary state is:

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Ionization energy of is 19.6×10−18 J atom−1. The energy of the first stationary state n = 1 of Li2+ is

The correct option is C 4.41 X 10-16 J atom-1IHe+ILi2+=Z21Z22 19.6×10−18ILi2+=49 ILi2+=94×19.6×10−18=44.1×10−18=4.41×10−17 J atom−1

Ionization energy of is $19.6 \times 10^{- 18} J a t o m^{- 1}$ . The energy of the first stationary state n = 1 of $L i^{2 +}$ is

The correct option is C 4.41 X 10^-16 J atom^-1
$\frac{I_{H e^{+}}}{I_{L i^{2 +}}} = \frac{Z_{1}^{2}}{Z_{2}^{2}}$
$\frac{19.6 \times 10^{- 18}}{I_{L i^{2 +}}} = \frac{4}{9}$
$I_{L i^{2 +}} = \frac{9}{4} \times 19.6 \times 10^{- 18} = 44.1 \times 10^{- 18} = 4.41 \times 10^{- 17} J a t o m^{- 1}$