The first ionization energies (removal of outermost electron) of cesium (Cs) and

Question

The first ionization energies (removal of outermost electron) of cesium (Cs) and gold (Au) are respectively 3.89 eV and 9.23 eV (1 eV/atom=96.487 kj/mol). Using the Bohr equation, calculate the ionization energy of an H atom with its one electron in the same orbital as the outermost electron in each of these two atoms. Account for the differences in balues (compare your value for H to those given for Cs and Au). Also, account for the difference between Cs and Au. Hint: you will need to determine the principle quantum number of the outermost electron.

Explanation / Answer

Answer: According to the Bohr model we know that the energy = -13.6 Z2/ n2 eV

here the values of Z is 1 and the principal quantum number is 1 and hence the ionization energy we get 13.6 ev for the hydrogen atom .

The value of ionization energy for H in ground level when principal quantum number is1 is higher then gold and cesium .

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