Question

The ionisation potential of a hydrogen atom is –13.6 eV. What will be the energy of the atom corresponding to n = 2.
(a) – 3.4 eV
(b) – 6.8 eV
(c) – 1.7 eV
(d) –2.7 eV

Answer 1

According to the question, the Ionisation potential of hydrogen atom is 13.6eV.

Therefore, the change in energy involved in removing the electron from n=2 is given by

the formula E(n)=−13.6n2 eV/atom

However, in this case the electron is completely removed from n=2, therefore

ΔE=E(∞)−E(2)

Therefore,

ΔE=0−(−13.622)

ΔE=+13.64

ΔE=3.4eV

So the answer is (a) -3.4eV