What is the number of revolutions that an electron will make per second in the `4^(th)` Bohr orbit of `Li^(+2)` ?

Options

A. `3.48 xx 10^(15) rps`
B. `4.55 xx 10^(13) rps`
C. `7.12 xx 10^(13) rps`
D. `9.22 xx 10^(14)` rps

Explanation

To find the number of revolutions that an electron will make per second in the 4th Bohr orbit of \( \text{Li}^{2+} \), we will follow these steps:

### Step 1: Determine the radius of the 4th Bohr orbit
The radius of the \( n^{th} \) Bohr orbit for a hydrogen-like atom is given by the formula:
\[ R_n = \frac{n^2 \cdot a_0}{Z} \]
where:
- \( n \) is the principal quantum number (4 in this case),
- \( a_0 \) is the Bohr radius (\( 5.29 \times 10^{-11} \, \text{m} \)),
- \( Z \) is the atomic number (for \( \text{Li}^{2+} \), \( Z = 3 \)).

Substituting the values:
\[ R_4 = \frac{4^2 \cdot 5.29 \times 10^{-11}}{3} = \frac{16 \cdot 5.29 \times 10^{-11}}{3} = \frac{84.64 \times 10^{-11}}{3} = 2.82 \times 10^{-11} \, \text{m} \]

### Step 2: Calculate the velocity of the electron in the 4th orbit
The velocity \( V_n \) of the electron in the \( n^{th} \) orbit is given by:
\[ V_n = \frac{Z \cdot e^2}{2 \cdot \epsilon_0 \cdot h} \cdot \frac{1}{n} \]
where:
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)),
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)).

Substituting the values for \( Z = 3 \) and \( n = 4 \):
\[ V_4 = \frac{3 \cdot (1.6 \times 10^{-19})^2}{2 \cdot (8.85 \times 10^{-12}) \cdot (6.63 \times 10^{-34})} \cdot \frac{1}{4} \]

Calculating \( V_4 \):
\[ V_4 = \frac{3 \cdot 2.56 \times 10^{-38}}{1.17 \times 10^{-45}} \cdot \frac{1}{4} = \frac{7.68 \times 10^{7}}{4} = 1.92 \times 10^{7} \, \text{m/s} \]

### Step 3: Calculate the frequency of revolution
The frequency \( f \) (number of revolutions per second) is given by:
\[ f = \frac{V_n}{2 \pi R_n} \]
Substituting the values for \( V_4 \) and \( R_4 \):
\[ f = \frac{1.92 \times 10^{7}}{2 \pi (2.82 \times 10^{-11})} \]

Calculating \( f \):
\[ f = \frac{1.92 \times 10^{7}}{1.77 \times 10^{-10}} \approx 1.08 \times 10^{17} \, \text{Hz} \]

### Final Answer
The number of revolutions that an electron will make per second in the 4th Bohr orbit of \( \text{Li}^{2+} \) is approximately \( 1.08 \times 10^{17} \, \text{Hz} \).

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