The first excitation energy of hydrogen atom is x eV. Then, the ionisation energy of the atom would be (in eV)
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To find the ionisation energy of a hydrogen like atom with one electron we use the formula
IE = 13.6Z²/n² eV
Where Z is atomic number and n is the shell number.
In this case Z is 3 and n is 1( because hydrogen is having n=1)
So IE Li²+ = 13.6×9 = 122.4 eV.
IE = 13.6Z²/n² eV
Where Z is atomic number and n is the shell number.
In this case Z is 3 and n is 1( because hydrogen is having n=1)
So IE Li²+ = 13.6×9 = 122.4 eV.
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