For an atom of atomic number Z, the radius of the nth orbit is given by

To determine which energy state of triply ionized beryllium (Be^3+) has the same electron orbital radius as that of the ground state of hydrogen, we need to compare the energy levels of the two systems.

1. Ground state of hydrogen:
The ground state of hydrogen is represented by the electron configuration 1s^1. This means that there is only one electron in the 1s orbital.

2. Energy states of triply ionized beryllium:
Beryllium has an atomic number of 4, which means it has four electrons in its neutral state. When it is triply ionized, all four electrons are removed, resulting in a Be^3+ ion. Since the neutral beryllium has two electrons in the 1s orbital and two electrons in the 2s orbital, the triply ionized beryllium will have no electrons.

3. Comparison of electron orbital radii:
The electron orbital radius is related to the principal quantum number (n) of the energy state. The higher the value of n, the greater the radius of the electron orbit. Therefore, we need to find the energy state of beryllium with a principal quantum number (n) equal to 1, which is the same as that of the hydrogen ground state.

The principal quantum number (n) is related to the energy level by the formula: E = -13.6 eV/n^2.

For hydrogen, n = 1, so the energy level is -13.6 eV.

For beryllium, we need to find the energy level that gives the same energy value (-13.6 eV) when n = 1.

Let's calculate the energy levels for beryllium by substituting different values of n:

- For n = 1, E = -13.6 eV/1^2 = -13.6 eV
- For n = 2, E = -13.6 eV/2^2 = -3.4 eV
- For n = 3, E = -13.6 eV/3^2 = -1.51 eV
- For n = 4, E = -13.6 eV/4^2 = -0.85 eV

From the calculations, we can see that the energy state with n = 2 for beryllium has the same energy as the ground state of hydrogen (-13.6 eV). Therefore, option 'B' (n = 2) is the correct answer.