Question

The angular momentum of an electron in a stationary state of \(Li^{2+}\) (\(Z=3\)) is \( \frac{3h}{\pi} \). The radius and energy of that stationary state are respectively.

• A. \(3.174 \text{ Å}, -5.45 \times 10^{-19} \text{ J}\)
• B. \(6.348 \text{ Å}, -5.45 \times 10^{-19} \text{ J}\)
• C. \(6.348 \text{ Å}, +5.45 \times 10^{-18} \text{ J}\)
• D. \(2.116 \text{ Å}, -5.45 \times 10^{-19} \text{ J}\)

Hint:

The radius of an orbit in a hydrogen-like atom follows \( r_n \propto \frac{n^2}{Z} \), while the energy follows \( E_n \propto -\frac{Z^2}{n^2} \). These formulas help in quick calculations.

Answer 1

The Correct Option is B

Step 1: Understanding the Given Data
We are given that the angular momentum of the electron in a stationary state of \(Li^{2+}\) (Z = 3) follows the quantization condition: \[ L = \frac{nh}{2\pi} \] Given \( L = \frac{3h}{\pi} \), we can write: \[ n = 3 \] Step 2: Formula for Bohr Radius
The Bohr radius for an ion with atomic number \(Z\) is given by: \[ r_n = \frac{n^2 a_0}{Z} \] where \( a_0 = 0.529 \text{ Å} \) is the Bohr radius. \[ r_3 = \frac{3^2 \times 0.529}{3} \] \[ = 6.348 \text{ Å} \] Step 3: Energy Calculation
The energy of the electron in the nth orbit is given by: \[ E_n = \frac{-13.6 Z^2}{n^2} \text{ eV} \] Substituting \( Z = 3 \), \( n = 3 \): \[ E_3 = \frac{-13.6 \times 9}{9} \] \[ = -13.6 \text{ eV} = -5.45 \times 10^{-19} \text{ J} \] Step 4: Conclusion
Thus, the correct values for the radius and energy are: \[ \boxed{6.348 \text{ Å}, -5.45 \times 10^{-19} \text{ J}} \]