Question

the ionization energy of hydrogen atom is 13.6 ev. the energy of the second excited state of He+ ion will be

Answer 1

Ionization energy can be calculated using the formula:

E=-13.6 \frac{z^{2}}{n^{2}}

z is the mass number, n is the number of orbit.

For a hydrogen atom , H:

z is 1 and n is 1 as it is for first excited state.

So energy comes out to be -13.6 J

And for a helium atom, He:

z is 2 and n is 2.

As here it need to be calculated for the second excited state.

So energy value comes out to be:

E= -13.6 \frac{2^{2}}{2^{2}}

E= -13.6 J.

Answer 2

The formula is :

En = - 13.6 Z² / n²

Z = 2

n = 2

Doing substitution :

E₂ = - 13.6 × (2²) (1/2² - 1/1²)

E₂ = 13.6 × 4 × (1 - 1/4)

E₂ = 13.6 × 4 × 3/4 = 40.8 eV

Answer :

40.8 eV