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, we need to determine which hydrogen-like atom or ion corresponds to the given ground state energy of -122.4 eV. We will use the formula for the energy of a hydrogen-like atom:

\[ E_n = -\frac{Z^2}{n^2} \times 13.6 \text{ eV} \]

where:
- \( E_n \) is the energy of the electron in the nth energy level,
- \( Z \) is the atomic number of the atom,
- \( n \) is the principal quantum number (for ground state, \( n = 1 \)),
- 13.6 eV is the energy of the ground state of hydrogen.

### Step 1: Set up the equation for the ground state energy.
Since we are given that the energy \( E \) is -122.4 eV and we are looking for the ground state (where \( n = 1 \)), we can write:

\[ -122.4 \text{ eV} = -\frac{Z^2}{1^2} \times 13.6 \text{ eV} \]

### Step 2: Simplify the equation.
Removing the negative signs from both sides gives us:

\[ 122.4 = Z^2 \times 13.6 \]

### Step 3: Solve for \( Z^2 \).
Now, we can isolate \( Z^2 \):

\[ Z^2 = \frac{122.4}{13.6} \]

### Step 4: Calculate \( Z^2 \).
Now we perform the division:

\[ Z^2 = \frac{122.4}{13.6} = 9 \]

### Step 5: Find \( Z \).
Taking the square root of both sides gives us:

\[ Z = \sqrt{9} = 3 \]

### Step 6: Identify the atom or ion.
The atomic number \( Z = 3 \) corresponds to lithium (Li). Since we are looking for a hydrogen-like ion, we consider the lithium ion with a +1 charge, which is \( \text{Li}^{2+} \).

### Conclusion:
The hydrogen-like atom or ion that has a ground state energy of -122.4 eV is \( \text{Li}^{2+} \).

### Final Answer:
The correct option is \( \text{Li}^{2+} \).

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