Calculate the velocity of an electron placed in the 2nd orbit of the `Li^(2+)` ion. Also calculate the number of revolutions per second that it around the nucleus.

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the velocity of the electron in the 2nd orbit of the \( \text{Li}^{2+} \) ion.

According to Bohr's model, the velocity \( v \) of an electron in the nth orbit is given by the formula:

\[ v = 2.18 \times 10^6 \times \frac{Z}{n} \text{ m/s} \]

Where:
- \( Z \) is the atomic number of the ion (for \( \text{Li}^{2+} \), \( Z = 3 \))
- \( n \) is the principal quantum number (for the 2nd orbit, \( n = 2 \))

Substituting the values:

\[ v = 2.18 \times 10^6 \times \frac{3}{2} \]

Calculating this:

\[ v = 2.18 \times 10^6 \times 1.5 = 3.27 \times 10^6 \text{ m/s} \]

### Step 2: Calculate the radius of the 2nd orbit.

The radius \( r \) of the nth orbit is given by:

\[ r = r_0 \times \frac{n^2}{Z} \]

Where:
- \( r_0 \) is the Bohr radius, approximately \( 0.5 \) Å (or \( 0.5 \times 10^{-10} \) m)
- \( n = 2 \)
- \( Z = 3 \)

Substituting the values:

\[ r = 0.5 \times 10^{-10} \times \frac{2^2}{3} \]

Calculating this:

\[ r = 0.5 \times 10^{-10} \times \frac{4}{3} = \frac{2}{3} \times 10^{-10} \approx 0.7053 \times 10^{-10} \text{ m} = 7.053 \times 10^{-11} \text{ m} \]

### Step 3: Calculate the number of revolutions per second.

The number of revolutions per second \( f \) can be calculated using the formula:

\[ f = \frac{v}{C} \]

Where \( C \) is the circumference of the orbit given by:

\[ C = 2 \pi r \]

Substituting for \( C \):

\[ C = 2 \pi \times 7.053 \times 10^{-11} \text{ m} \]

Calculating \( C \):

\[ C \approx 4.43 \times 10^{-10} \text{ m} \]

Now substituting \( v \) and \( C \) into the formula for \( f \):

\[ f = \frac{3.27 \times 10^6}{4.43 \times 10^{-10}} \]

Calculating \( f \):

\[ f \approx 7.38 \times 10^{15} \text{ revolutions per second} \]

### Final Answers:
1. The velocity of the electron in the 2nd orbit of the \( \text{Li}^{2+} \) ion is \( 3.27 \times 10^6 \text{ m/s} \).
2. The number of revolutions per second is \( 7.38 \times 10^{15} \text{ revolutions/second} \).