In Bohr's atomic model, the energy of an electron in an atom is given by the formula:

$ E \propto \frac{Z^2}{n^2} $

where $ Z $ is the atomic number and $ n $ is the principal quantum number of the electron's orbit. Let’s consider the implications:

For hydrogen ($ \mathrm{H} $), $ Z = 1 $.

For helium ion ($ \mathrm{He}^{+} $), $ Z = 2 $.

For lithium ion ($ \mathrm{Li}^{++} $), $ Z = 3 $.

Now, identifying the relevant states:

Ground State: This corresponds to $ n = 1 $.

First Excited State: This corresponds to $ n = 2 $.

Second Excited State: This corresponds to $ n = 3 $.

Let's evaluate the energy comparisons:

Hydrogen Atom in Ground State: $ E \propto \frac{1^2}{1^2} = 1 $.

Helium Ion ($ \mathrm{He}^{+} $) in the First Excited State:

$ E \propto \frac{2^2}{2^2} = 1 $

This matches the energy of hydrogen in its ground state.

Lithium Ion ($ \mathrm{Li}^{++} $) in the Second Excited State:

$ E \propto \frac{3^2}{3^2} = 1 $

This also matches the energy of hydrogen in its ground state.

From this analysis, we can conclude:

The energy of a hydrogen atom in its ground state is equal to the energy of a $ \mathrm{He}^{+} $ ion in its first excited state.

The energy of a hydrogen atom in its ground state also equals the energy of a $ \mathrm{Li}^{++} $ ion in its second excited state.

Hence, statements (A) and (B) are correct.