Question

Q13. A hydrogen atom can be exited to higher state by supplying external energy. A photon
of energy 9.2 eV is allowed to fall on the hydrogen atom. Will hydrogen atom be excited?
Explain.​

Answer 1

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 or less than it.

Explanation:

hope is helpful