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The ionisation Energy of He^+ ion is 19...

The ionisation Energy of `He^+ ion` is `19.6x10^-18 J atom^1`. Find the energy of the last orbit of

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The energy of the first orbit of Li^(+2) = ((-19.6xx10^-18xx9)/4) J atom^-1 = -4.41xx10^-17 J atom^-1 (E_npropz^2/n^2)`
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Knowledge Check

  • Ionisation energy of He^+ is 19.6xx10^(-18) J atom^(-1) The energy of the first stationary state (n=1) of Li^(2+)?

    A
    `4.41xx10^(-16) Jatom^(-1)`
    B
    `-4.41xx10^(-17) Jatom^(-1)`
    C
    `-2.2xx10^(-15) Jatom^(-1)`
    D
    `8.82xx10^(-17) Jatom^(-1)`

  • If the ionization energy of Li^(+2) is 19.6xx10^(-18) J per atom then energy of Be^(+3) ion in the second stationary state is

    A
    `-4.9xx10^-18J`
    B
    `-19.6xx10^-18J`
    C
    `-11.025xx10^-18J`
    D
    none of these

  • It the 1st ionisation energy of H atom is 13.6 eV, then the 2nd ionisation energy of the atom is-

    A
    27.2 eV
    B
    40.8 eV
    C
    54.4 eV
    D
    108.8 eV

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    The ionisation energy for the H-atom in the ground state is xJ*"atom"^(-1) . Show that the energy requried for the process, He^(+)(g) to He^(2+)+e , is 4x" "J*"atom"^(-1) .

    If the 1st ionisation energy of H atom is 13.6 eV then the 2nd ionisation energy of He atom is

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