Which energy state of triply ionized beryllium `(Be^(+3))` has the same orbital radius as that of state of hydrogen atom

Options

A. n = 8 state
B. n = 5 state
C. n = 4 state
D. n = 2 state.

Explanation

To solve the problem of finding which energy state of triply ionized beryllium \((Be^{+3})\) has the same orbital radius as that of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Orbital Radius**:
The formula for the radius of the \(n^{th}\) orbit in a hydrogen-like atom is given by:
\[ R_n = 0.529 \frac{n^2}{Z} \text{ Å} \]
where \(R_n\) is the radius, \(n\) is the principal quantum number, and \(Z\) is the atomic number.

2. **Calculate the Radius for Hydrogen**:
For hydrogen, \(Z = 1\) and the principal quantum number \(n = 1\):
\[ R_H = 0.529 \frac{1^2}{1} = 0.529 \text{ Å} \]

3. **Set Up the Equation for Beryllium**:
For triply ionized beryllium \((Be^{+3})\), the atomic number \(Z = 4\). We need to find the value of \(n\) such that the radius of beryllium equals the radius of hydrogen:
\[ R_{Be} = 0.529 \frac{n^2}{4} \]

4. **Equate the Radii**:
Since we want \(R_{Be} = R_H\):
\[ 0.529 \frac{n^2}{4} = 0.529 \]

5. **Simplify the Equation**:
We can cancel \(0.529\) from both sides:
\[ \frac{n^2}{4} = 1 \]

6. **Solve for \(n^2\)**:
Multiply both sides by 4:
\[ n^2 = 4 \]

7. **Find \(n\)**:
Taking the square root of both sides gives:
\[ n = 2 \]

8. **Conclusion**:
Therefore, the energy state of triply ionized beryllium \((Be^{+3})\) that has the same orbital radius as that of the hydrogen atom is \(n = 2\).

### Final Answer:
The energy state of triply ionized beryllium \((Be^{+3})\) that has the same orbital radius as that of hydrogen is \(n = 2\).

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