If `a_(0)` is the Bohr radius, the radius of then `n=2` electronic orbit in triply ionized beryllium is

Options

A. `4a_(0)`
B. `a_(0)`
C. `a_(0)//4`
D. `a_(0)//16`

Explanation

To find the radius of the n=2 electronic orbit in triply ionized beryllium (Be³⁺), we can use the formula for the radius of an electron orbit in a hydrogen-like atom:

\[ r_n = \frac{n^2 a_0}{Z} \]

where:
- \( r_n \) is the radius of the orbit,
- \( n \) is the principal quantum number (in this case, \( n = 2 \)),
- \( a_0 \) is the Bohr radius (approximately \( 5.29 \times 10^{-11} \) m),
- \( Z \) is the atomic number of the ion.

### Step 1: Identify the values
For triply ionized beryllium (Be³⁺):
- The atomic number \( Z \) of beryllium is 4.
- The principal quantum number \( n = 2 \).

### Step 2: Substitute the values into the formula
Now, we can substitute these values into the formula:

\[ r_2 = \frac{n^2 a_0}{Z} \]

Substituting \( n = 2 \) and \( Z = 4 \):

\[ r_2 = \frac{2^2 a_0}{4} \]

### Step 3: Calculate \( r_2 \)
Calculating \( 2^2 \):

\[ 2^2 = 4 \]

Now substituting this back into the equation:

\[ r_2 = \frac{4 a_0}{4} \]

This simplifies to:

\[ r_2 = a_0 \]

### Conclusion
Thus, the radius of the n=2 electronic orbit in triply ionized beryllium is:

\[ r_2 = a_0 \]