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Question: The ionization energy of \(He^{+}\) is \(19.6 \times 10^{- 18}\)Jatom<sup>–1</sup>. Calculate the en...

The ionization energy of He+He^{+} is 19.6×10−1819.6 \times 10^{- 18}Jatom–1. Calculate the energy of the first stationary state of Li+2Li^{+ 2}

A

19.6×10−18Jatom-119.6 \times 10^{- 18}J\text{ato}\text{m}^{\text{-1}}

B

4.41×10−18J4.41 \times 10^{- 18}J atom–1

C

19.6×10−19Jatom-119.6 \times 10^{- 19}J\text{ato}\text{m}^{\text{-1}}

D

4.41×10−17Jatom−14.41 \times 10^{- 17}J\text{ato}\text{m}^{- 1}

Answer

4.41×10−17Jatom−14.41 \times 10^{- 17}J\text{ato}\text{m}^{- 1}

Explanation

Solution

I.E. of He+=E×22(ZHe^{+} = E \times 2^{2}(Z for He=2)He = 2)

I.E. of Li2+=E×33Li^{2 +} = E \times 3^{3} (Z for Li=3)

∴I.E.(He+)I.E.(Li2+)=49\therefore\frac{I.E.(He^{+})}{I.E.(Li^{2 +})} = \frac{4}{9} or I.E. (Li2+)=94×I.E.(He+)(Li^{2 +}) = \frac{9}{4} \times I.E.(He^{+})

=94×19.6×10−18= \frac{9}{4} \times 19.6 \times 10^{- 18} =4.41×10−174.41 \times 10^{- 17} J atom–1