Which state of the triply ionized Beryllium `(Be^(3+))` has the same orbit radius as that of the ground state of hydrogen atom?
A
3
B
2
C
4
D
5
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem of determining which state of the triply ionized Beryllium `(Be^(3+))` has the same orbit radius as that of the ground state of a hydrogen atom, we will use Bohr's model of the atom.
### Step-by-Step Solution:
1. **Understand the Bohr Model Formula**:
The radius of an electron in a hydrogen-like atom is given by the formula:
\[
r = a_0 \frac{n^2}{Z}
\]
where:
- \( r \) is the radius of the orbit,
- \( a_0 \) is the Bohr radius (approximately \( 0.529 \, \text{Å} \)),
- \( n \) is the principal quantum number (orbit number),
- \( Z \) is the atomic number of the element.
2. **Identify Parameters for Hydrogen**:
For the hydrogen atom:
- \( Z = 1 \) (since hydrogen has one proton),
- The ground state corresponds to \( n = 1 \).
Thus, the radius of the ground state of hydrogen is:
\[
r_H = a_0 \frac{1^2}{1} = a_0
\]
3. **Identify Parameters for Triply Ionized Beryllium**:
For triply ionized Beryllium `(Be^(3+))`:
- The atomic number \( Z = 4 \) (Beryllium has four protons),
- We need to find the value of \( n \) such that the radius of the orbit equals that of hydrogen.
4. **Set Up the Equation for Beryllium**:
We want the radius of the orbit in Beryllium to equal that of hydrogen:
\[
r_{Be^{3+}} = a_0 \frac{n^2}{Z} = a_0
\]
Substituting \( Z = 4 \):
\[
a_0 \frac{n^2}{4} = a_0
\]
5. **Solve for \( n^2 \)**:
Dividing both sides by \( a_0 \) (assuming \( a_0 \neq 0 \)):
\[
\frac{n^2}{4} = 1
\]
Multiplying both sides by 4:
\[
n^2 = 4
\]
6. **Find \( n \)**:
Taking the square root of both sides:
\[
n = 2
\]
### Conclusion:
The state of the triply ionized Beryllium `(Be^(3+))` that has the same orbit radius as that of the ground state of hydrogen atom is when \( n = 2 \), which corresponds to the first excited state.
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