What is the frequency of revolution of electron present in `2nd` Bohr's orbit of `H-` atom ?

Options

A. `1.016xx10^(16)s^(-1)`
B. `4.065xx10^(16)s^(-1)`
C. `1.626xx10^(15)s^(-1)`
D. `8.13xx10^(16)s^(-1)`

Explanation

To find the frequency of revolution of the electron present in the 2nd Bohr's orbit of a hydrogen atom (H-), we will follow these steps:

### Step 1: Calculate the radius of the 2nd Bohr orbit (r2)
The formula for the radius of the nth Bohr orbit is given by:

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

For hydrogen (Z = 1) and n = 2:

\[ r_2 = \frac{0.529 \, \text{Å} \cdot 2^2}{1} = 0.529 \times 4 \, \text{Å} = 2.116 \, \text{Å} \]

Converting Ångströms to meters (1 Å = \(10^{-10}\) m):

\[ r_2 = 2.116 \times 10^{-10} \, \text{m} \]

### Step 2: Calculate the velocity of the electron in the 2nd Bohr orbit (v2)
The velocity of the electron in the nth orbit is given by:

\[ v_n = \frac{2.2 \times 10^6 \cdot Z}{n} \, \text{m/s} \]

For hydrogen (Z = 1) and n = 2:

\[ v_2 = \frac{2.2 \times 10^6 \cdot 1}{2} = 1.1 \times 10^6 \, \text{m/s} \]

### Step 3: Calculate the circumference of the 2nd Bohr orbit (C)
The circumference (C) of the orbit is given by:

\[ C = 2 \pi r \]

Substituting the value of r2:

\[ C = 2 \pi (2.116 \times 10^{-10}) \approx 2 \times 3.14 \times 2.116 \times 10^{-10} \approx 1.33 \times 10^{-9} \, \text{m} \]

### Step 4: Calculate the frequency of revolution (f)
The frequency of revolution is given by:

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

Substituting the values of v2 and C:

\[ f = \frac{1.1 \times 10^6}{1.33 \times 10^{-9}} \approx 8.27 \times 10^{14} \, \text{s}^{-1} \]

### Final Answer
The frequency of revolution of the electron in the 2nd Bohr orbit of the hydrogen atom is approximately:

\[ f \approx 8.27 \times 10^{14} \, \text{s}^{-1} \]