The number of revolutions per second traced by an electron in the 2nd bohr orbit of `Li^(+2)` will be

Options

A. `6.15 xx 10^(16)` rps
B. `7.37 xx 10^(15)` rps
C. `1.53 xx 10^(11)` rps
D. `8.92 xx 10^(16)` rps

Answer

To find the number of revolutions per second traced by an electron in the 2nd Bohr orbit of \( \text{Li}^{2+} \), we can follow these steps:

### Step 1: Determine the radius of the 2nd Bohr orbit

The formula for the radius of the nth orbit in a hydrogen-like atom is given by:

\[ r_n = \frac{0.529 \, n^2}{Z} \, \text{Å} \]

Where:
- \( n \) = principal quantum number (for the 2nd orbit, \( n = 2 \))
- \( Z \) = atomic number (for lithium, \( Z = 3 \))

Substituting the values:

\[ r_2 = \frac{0.529 \times 2^2}{3} = \frac{0.529 \times 4}{3} = \frac{2.116}{3} \approx 0.7053 \, \text{Å} \]

### Step 2: Convert the radius to meters

Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \):

\[ r_2 = 0.7053 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the velocity of the electron in the 2nd orbit

The velocity for the nth orbit is given by:

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

Substituting \( Z = 3 \) and \( n = 2 \):

\[ v_2 = 2.18 \times 10^6 \frac{3}{2} = 3.27 \times 10^6 \, \text{m/s} \]

### Step 4: Calculate the frequency (number of revolutions per second)

The frequency \( f \) can be calculated using the formula:

\[ f = \frac{v_n}{2 \pi r_n} \]

Substituting the values we have:

\[ f = \frac{3.27 \times 10^6}{2 \times 3.14 \times (0.7053 \times 10^{-10})} \]

Calculating the denominator:

\[ 2 \times 3.14 \times 0.7053 \times 10^{-10} \approx 4.43 \times 10^{-10} \]

Now substituting this back into the frequency equation:

\[ f = \frac{3.27 \times 10^6}{4.43 \times 10^{-10}} \approx 7.37 \times 10^{15} \, \text{revolutions per second} \]

### Final Answer

Thus, the number of revolutions per second traced by an electron in the 2nd Bohr orbit of \( \text{Li}^{2+} \) is approximately:

\[ \boxed{7.37 \times 10^{15} \, \text{RPS}} \]

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