What is the energy of state in a triply ionized beryllium `(Be^(+++))` whose radius is equal to that of ground state of hydrogen ?

Options

A. 27.2eV
B. 54.4eV
C. 13.6eV
D. 40.8eV

Answer

To find the energy of the triply ionized beryllium ion (Be^(+++)) whose radius is equal to that of the ground state of hydrogen, we can follow these steps:

### Step 1: Understand the relationship between radius and quantum numbers
The radius of an electron in a hydrogen-like atom is given by the formula:
\[ r = r_0 \frac{Z}{n^2} \]
where:
- \( r_0 \) is the Bohr radius (the radius of the ground state of hydrogen),
- \( Z \) is the atomic number,
- \( n \) is the principal quantum number.

### Step 2: Set the radius of Be^(+++) equal to the radius of hydrogen
Since we are given that the radius of Be^(+++) is equal to that of the ground state of hydrogen, we can write:
\[ r_{Be^{+++}} = r_{H} \]
This implies:
\[ r_0 \frac{Z_{Be^{+++}}}{n^2} = r_0 \]
Since \( r_0 \) cancels out, we get:
\[ \frac{Z_{Be^{+++}}}{n^2} = 1 \]

### Step 3: Identify the atomic number of Be^(+++)
The atomic number \( Z \) for beryllium is 4. Therefore, we have:
\[ \frac{4}{n^2} = 1 \]
From this, we can solve for \( n^2 \):
\[ n^2 = 4 \]
Thus, \( n = 2 \).

### Step 4: Use the energy formula for hydrogen-like atoms
The energy of an electron in a hydrogen-like atom is given by the formula:
\[ E = -\frac{13.6 Z^2}{n^2} \]
Substituting \( Z = 4 \) and \( n = 2 \):
\[ E = -\frac{13.6 \times 4^2}{2^2} \]
\[ E = -\frac{13.6 \times 16}{4} \]
\[ E = -\frac{217.6}{4} \]
\[ E = -54.4 \text{ eV} \]

### Step 5: Conclusion
The energy of the triply ionized beryllium ion (Be^(+++)) whose radius is equal to that of the ground state of hydrogen is:
\[ E = -54.4 \text{ eV} \]