Which state of triply ionised Beryllium `(Be^(+++))` the same orbital radius as that of the ground state hydrogen ?

Options

A. `n = 4`
B. `n = 3`
C. `n = 2`
D. `n = 1`

Explanation

To find which state of triply ionized Beryllium `(Be^(+++))` has the same orbital radius as that of the ground state hydrogen, we can follow these steps:

### Step 1: Understand the formula for orbital radius
The formula for the radius of an electron in an atom is given by:
\[ r_n = \frac{r_0 \cdot n^2}{Z} \]
where:
- \( r_n \) is the radius of the nth orbit,
- \( r_0 \) is the Bohr radius (approximately \( 5.29 \times 10^{-11} \) m),
- \( n \) is the principal quantum number (1 for ground state),
- \( Z \) is the atomic number of the element.

### Step 2: Calculate the radius of the ground state hydrogen
For hydrogen, which has an atomic number \( Z = 1 \):
\[ r_1 = \frac{r_0 \cdot 1^2}{1} = r_0 \]
Thus, the radius of the ground state of hydrogen is simply \( r_0 \).

### Step 3: Determine the atomic number of triply ionized Beryllium
Triply ionized Beryllium `(Be^(+++))` has lost three electrons, leaving it with one electron. The atomic number of Beryllium is \( Z = 4 \).

### Step 4: Set up the equation for Beryllium
We want to find the principal quantum number \( n \) for Beryllium such that its radius equals that of hydrogen's ground state:
\[ r_n = \frac{r_0 \cdot n^2}{Z} \]
Setting this equal to \( r_0 \):
\[ \frac{r_0 \cdot n^2}{4} = r_0 \]

### Step 5: Solve for \( n \)
We can simplify the equation by dividing both sides by \( r_0 \):
\[ \frac{n^2}{4} = 1 \]
Multiplying both sides by 4 gives:
\[ n^2 = 4 \]
Taking the square root of both sides:
\[ n = 2 \]

### Conclusion
The state of triply ionized Beryllium `(Be^(+++))` that has the same orbital radius as that of the ground state hydrogen is the second orbital, where \( n = 2 \).