calculate the ionization energy of the hydrogen atom in its ground state in electron volt
Answers
Answer:
The ionization energy (IE) of an atom describes the minimum amount of energy required to remove an electron (to infinity) from the atom.
For the electron in ground state i.e. n = 1, the ionization energy is
IE
1
=E
∞
−E
1
IE
1
=0−(−13.6) .................(since, ground state energy is (-13.6 eV))
IE
1
=13.6eV
For the electron in first excited state i.e. n = 2, the ionization energy is
IE
2
=
n
2
IE
1
IE
2
=
2
2
IE
1
IE
2
=
4
13.6
=3.4eV
For the electron in second excited state i.e. n = 3, the ionization energy is
IE
2
=
n
2
IE
1
IE
2
=
3
2
IE
1
IE
2
=
9
13.6
=1.51eV and so on.
Hence, the ionisation energy for excited hydrogen atom will be 3.4 eV
Explanation: