For a doubly ionised Li-atom

Options

A. Angular momentum of an electron in `3rd` orbit is `(3h)/(2pi)`
B. energy of electron in `2nd` ecited state is `-13.6 eV`.
C. Speed of electron in `3rd` orbit is `(c)/(137)`, where c is speed of light
D. Kinetic energy of electron is 2nd excited state is half of the magnitude of the potential energy.

Explanation

To solve the problem regarding the angular momentum of an electron in a doubly ionized lithium atom (Li²⁺) in the third orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- A doubly ionized lithium atom (Li²⁺) has lost two electrons, leaving one electron in its outer shell. The atomic number (Z) of lithium is 3.

2. **Using Bohr's Model**:
- According to Bohr's model of the atom, the angular momentum (L) of an electron in a given orbit is quantized and is given by the formula:
\[ L = n \frac{h}{2\pi} \]
- Here, \( n \) is the principal quantum number (the orbit number), and \( h \) is Planck's constant.

3. **Identifying the Orbit**:
- For the third orbit, \( n = 3 \).

4. **Calculating Angular Momentum**:
- Substituting \( n \) into the angular momentum formula:
\[ L = 3 \frac{h}{2\pi} \]
- This confirms that the angular momentum of the electron in the third orbit is indeed \( \frac{3h}{2\pi} \).

5. **Energy Levels**:
- The energy of an electron in a hydrogen-like atom is given by:
\[ E_n = -\frac{Z^2 \cdot 13.6 \text{ eV}}{n^2} \]
- For Li²⁺, \( Z = 3 \) and for the second excited state (which corresponds to \( n = 3 \)):
\[ E_3 = -\frac{3^2 \cdot 13.6 \text{ eV}}{3^2} = -13.6 \text{ eV} \]

6. **Kinetic Energy**:
- The kinetic energy (KE) of the electron can be calculated as:
\[ KE = -\frac{E_n}{2} = \frac{13.6 \text{ eV}}{2} = 6.8 \text{ eV} \]

7. **Conclusion**:
- The angular momentum for the electron in the third orbit of a doubly ionized lithium atom is \( \frac{3h}{2\pi} \), and the energy in the second excited state is \( -13.6 \text{ eV} \).