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. `8.2xx10^(14)s^(-1)`

Answer

To find the frequency of revolution of the electron present in the second Bohr's orbit of the hydrogen atom, we can follow these steps:

### Step 1: Calculate the radius of the second Bohr's orbit
According to Bohr's theory, the radius of the nth orbit is given by the formula:
\[ r_n = 0.529 \frac{n^2}{z} \]
where:
- \( n \) = principal quantum number (2 for the second orbit)
- \( z \) = atomic number (1 for hydrogen)

For the second Bohr's orbit:
\[ r_2 = 0.529 \frac{2^2}{1} = 0.529 \times 4 = 2.116 \, \text{Å} \]

### Step 2: Convert the radius from angstroms to meters
1 angstrom (Å) = \( 10^{-10} \) meters, so:
\[ r_2 = 2.116 \, \text{Å} = 2.116 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the velocity of the electron in the second orbit
The velocity \( v \) of the electron can be calculated using the formula:
\[ v = 2.2 \times 10^6 \frac{z}{n} \, \text{m/s} \]

For hydrogen (\( z = 1 \)) and the second orbit (\( n = 2 \)):
\[ v = 2.2 \times 10^6 \frac{1}{2} = 1.1 \times 10^6 \, \text{m/s} \]

### Step 4: Calculate the frequency of revolution
The frequency \( f \) of revolution is given by:
\[ f = \frac{v}{2 \pi r} \]

Substituting the values:
\[ f = \frac{1.1 \times 10^6}{2 \pi (2.116 \times 10^{-10})} \]

Calculating the denominator:
\[ 2 \pi (2.116 \times 10^{-10}) \approx 1.327 \times 10^{-9} \]

Now substituting this back into the frequency formula:
\[ f = \frac{1.1 \times 10^6}{1.327 \times 10^{-9}} \approx 8.29 \times 10^{14} \, \text{s}^{-1} \]

### Final Answer
The frequency of revolution of the electron present in the second Bohr's orbit of the hydrogen atom is approximately:
\[ f \approx 8.2 \times 10^{14} \, \text{s}^{-1} \]

---