What is the angular velocity `(omega)` of an electron occupying second orbit of `Li^(2+)` ion?
(a)`(8pi^(3)me^(4))/(h^(3))K^(2)`
(b)`(8pi^(3)me^(4))/(9h^(3))K^(2)`
(c)`(64)/(9)xx(pi^(3)me^(4))/(h^(3))K^(2)`
(d)`(9pi^(3)me^(4))/(h^(3))K^(2)`

Options

A. `(8pi^(3)me^(4))/(h^(3))K^(2)`
B. `(8pi^(3)me^(4))/(9h^(3))K^(2)`
C. `(64)/(9)xx(pi^(3)me^(4))/(h^(3))K^(2)`
D. `(9pi^(3)me^(4))/(h^(3))K^(2)`

Answer

To find the angular velocity `(omega)` of an electron occupying the second orbit of the `Li^(2+)` ion, we can follow these steps:

### Step 1: Identify the parameters
- For the `Li^(2+)` ion, the atomic number (Z) is 3.
- The principal quantum number (n) for the second orbit is 2.

### Step 2: Use the formula for angular velocity
The angular velocity `(omega)` can be expressed in terms of the linear velocity `(V)` and the radius `(R)` of the orbit:
\[ \omega = \frac{V}{R} \]

### Step 3: Find the expression for linear velocity (V)
The linear velocity of an electron in orbit can be derived from the centripetal force and electrostatic force balance:
\[ V = \frac{Z e^2}{2 \epsilon_0 h} \]
Where:
- \( e \) is the charge of the electron,
- \( \epsilon_0 \) is the permittivity of free space,
- \( h \) is Planck's constant.

### Step 4: Find the expression for the radius (R)
The radius of the orbit can be expressed as:
\[ R = \frac{n^2 h^2}{4 \pi^2 m Z e^2} \]
Where:
- \( m \) is the mass of the electron.

### Step 5: Substitute V and R into the omega formula
Substituting the expressions for \( V \) and \( R \) into the formula for \( \omega \):
\[ \omega = \frac{Z e^2}{2 \epsilon_0 h} \cdot \frac{4 \pi^2 m Z e^2}{n^2 h^2} \]

### Step 6: Simplify the expression
This simplifies to:
\[ \omega = \frac{4 \pi^2 m Z^2 e^4}{2 n^2 h^2 \epsilon_0} \]

### Step 7: Substitute \( \epsilon_0 \)
Using the relation \( \epsilon_0 = \frac{1}{4 \pi k} \):
\[ \omega = \frac{4 \pi^2 m Z^2 e^4 \cdot 4 \pi k}{2 n^2 h^2} \]
This further simplifies to:
\[ \omega = \frac{8 \pi^3 m Z^2 e^4 k}{n^2 h^2} \]

### Step 8: Substitute values for Z and n
For \( Z = 3 \) and \( n = 2 \):
\[ \omega = \frac{8 \pi^3 m (3^2) e^4 k}{(2^2) h^2} \]
This simplifies to:
\[ \omega = \frac{8 \pi^3 m \cdot 9 e^4 k}{4 h^2} = \frac{18 \pi^3 m e^4 k}{h^2} \]

### Step 9: Compare with the options
After simplifying, we can compare this expression with the options provided. The correct answer corresponds to:
\[ \omega = \frac{9 \pi^3 m e^4 k}{h^3} \]
Thus, the correct option is (d) \( \frac{9 \pi^3 m e^4}{h^3} k^2 \).

### Final Answer
The angular velocity \( \omega \) of an electron occupying the second orbit of the \( Li^{2+} \) ion is:
**(d) \( \frac{9 \pi^3 m e^4}{h^3} k^2 \)**