The energy of a hydrogen like atom (or ion) in its ground state is -122.4 eV. It may be

Options

A. hydrogen atom
B. `He^+`
C. `Li^(2+)`
D. `Be^(3+)`

Explanation

To solve the problem of identifying the hydrogen-like atom or ion with a ground state energy of -122.4 eV, we can follow these steps:

### Step 1: Understand the Energy Formula
The energy of a hydrogen-like atom in its nth energy level is given by the formula:
\[ E_n = -\frac{Z^2}{n^2} \times 13.6 \text{ eV} \]
where \(E_n\) is the energy at level \(n\), \(Z\) is the atomic number, and \(n\) is the principal quantum number.

### Step 2: Identify the Ground State
Since we are dealing with the ground state of the atom, we set \(n = 1\).

### Step 3: Substitute the Known Values
We know that the energy \(E_n\) is -122.4 eV. Substituting the known values into the formula:
\[ -122.4 = -\frac{Z^2}{1^2} \times 13.6 \]
This simplifies to:
\[ -122.4 = -Z^2 \times 13.6 \]

### Step 4: Solve for \(Z^2\)
Removing the negative signs and rearranging the equation gives:
\[ Z^2 = \frac{122.4}{13.6} \]

### Step 5: Calculate \(Z^2\)
Now, perform the division:
\[ Z^2 = \frac{122.4}{13.6} = 9 \]

### Step 6: Find \(Z\)
Taking the square root of both sides:
\[ Z = \sqrt{9} = 3 \]

### Step 7: Identify the Atom or Ion
The atomic number \(Z = 3\) corresponds to lithium (Li). Since we are looking for a hydrogen-like ion, we specifically have \(Li^{2+}\) (lithium with a +2 charge).

### Conclusion
Thus, the hydrogen-like atom or ion with a ground state energy of -122.4 eV is \(Li^{2+}\).

### Final Answer
The correct option is **helium two plus (Li^{2+})**.

---