The energy of a hydrogen atom in its ground state is -122.4ev. It may be a.Hydrogen atom b.He c.Li.2+ d.Be3+

To determine which atom corresponds to the given energy of -122.4 eV in its ground state, we need to consider the energy levels of hydrogen-like atoms. Hydrogen-like atoms are those that have only one electron, similar to hydrogen. This includes ions like He⁺, Li²⁺, and Be³⁺. The energy of an electron in a hydrogen-like atom can be calculated using the formula:

Energy Level Formula

The energy levels of a hydrogen-like atom are given by the equation:

E = -Z² * 13.6 eV / n²

Where:

• E is the energy of the electron.
• Z is the atomic number of the element.
• n is the principal quantum number (for ground state, n = 1).

Calculating for Each Atom

Let's calculate the energy for each option using the formula:

1. Hydrogen (H)

For hydrogen, Z = 1:

E = -1² * 13.6 eV / 1² = -13.6 eV

2. Helium Ion (He⁺)

For He⁺, Z = 2:

E = -2² * 13.6 eV / 1² = -54.4 eV

3. Lithium Ion (Li²⁺)

For Li²⁺, Z = 3:

E = -3² * 13.6 eV / 1² = -122.4 eV

4. Beryllium Ion (Be³⁺)

For Be³⁺, Z = 4:

E = -4² * 13.6 eV / 1² = -217.6 eV

Identifying the Correct Atom

From our calculations, we see that the energy of -122.4 eV corresponds to the lithium ion (Li²⁺). This is because it has three protons and, when stripped of its two electrons, behaves like a hydrogen-like atom with Z = 3.

Summary

In summary, the energy of -122.4 eV in its ground state is characteristic of the lithium ion (Li²⁺). This illustrates how the energy levels of hydrogen-like atoms depend on their atomic number, with higher atomic numbers resulting in more negative energy values due to the increased nuclear charge attracting the single electron more strongly.