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Question: If the first ionisation energy of hydrogen is \(2.179 \times 10^{- 18}J\) per atom, the second ionis...

If the first ionisation energy of hydrogen is 2.179×10−18J2.179 \times 10^{- 18}J per atom, the second ionisation energy of helium per atom is

A

8.716×10−18J8.716 \times 10^{- 18}J

B

5.5250kJ5.5250kJ

C

7.616×10−18J7.616 \times 10^{- 18}J

D

8.016×10−13J8.016 \times 10^{- 13}J

Answer

8.716×10−18J8.716 \times 10^{- 18}J

Explanation

Solution

For Bohrs systems : energy of the electron ∝Z2n2\propto \frac{Z^{2}}{n^{2}}

Ionisation energy is the difference of energies of an electron (E∞),(E_{\infty}), when taken to infinite distance and ErE_{r} when present in any Bohr orbit and EαE_{\alpha} is taken as zero so ionisation energy becomes equal to the energy of electron in any Bohr orbit.

EH∝ZH2nH2E_{H} \propto \frac{Z_{H}^{2}}{n_{H}^{2}} ; EHe∝ZHe2nHe2E_{He} \propto \frac{Z_{He}^{2}}{n_{He}^{2}} or EHEHe=12×2\frac{E_{H}}{E_{He}} = \frac{1}{2 \times 2}

[as ZH=1,ZHe=2,nH=1,nHe=1]Z_{H} = 1,Z_{He} = 2,n_{H} = 1,n_{He} = 1\rbrack

orEHe=EH×4=2.179×10−18×4=8.716×10−18E_{He} = E_{H} \times 4 = 2.179 \times 10^{- 18} \times 4 = 8.716 \times 10^{- 18}

Joule per atom.