An electron is moving in 3rd orbit of Hydrogen atom . The frequency of moving electron is

Options

A. `2.19 xx 10^(14)rps`
B. `7.3 xx 10^(14)` rps
C. `2.44 xx 10^(14)` rps
D. `7.3 xx 10^(10) ` rps

Answer

To find the frequency of an electron moving in the 3rd orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the relationship between frequency and time period
The frequency (ν) of an electron is the reciprocal of the time period (T). This can be expressed mathematically as:
\[ \nu = \frac{1}{T} \]

### Step 2: Use the formula for the time period
The time period for an electron in the nth orbit of a hydrogen atom can be calculated using the formula:
\[ T = n^3 \cdot \frac{4 \pi \epsilon_0 h^3}{m e^2} \]
where:
- \( n \) = principal quantum number (for the 3rd orbit, \( n = 3 \))
- \( \epsilon_0 \) = permittivity of free space \( = 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( h \) = Planck's constant \( = 6.626 \times 10^{-34} \, \text{Js} \)
- \( m \) = mass of the electron \( = 9.1 \times 10^{-31} \, \text{kg} \)
- \( e \) = charge of the electron \( = 1.6 \times 10^{-19} \, \text{C} \)

### Step 3: Substitute the values into the formula
Substituting \( n = 3 \) into the time period formula:
\[ T = 3^3 \cdot \frac{4 \pi \epsilon_0 h^3}{m e^2} \]
Calculating \( 3^3 \):
\[ 3^3 = 27 \]
Thus, we have:
\[ T = 27 \cdot \frac{4 \pi (8.85 \times 10^{-12}) (6.626 \times 10^{-34})^3}{(9.1 \times 10^{-31}) (1.6 \times 10^{-19})^2} \]

### Step 4: Calculate the value of T
Now, we will compute the value of \( T \):
1. Calculate \( h^3 \):
\[ h^3 = (6.626 \times 10^{-34})^3 \]
2. Calculate \( e^2 \):
\[ e^2 = (1.6 \times 10^{-19})^2 \]
3. Substitute these values into the equation for \( T \) and compute.

### Step 5: Find the frequency
Once we have \( T \), we can find the frequency using:
\[ \nu = \frac{1}{T} \]
After calculating \( T \), take the reciprocal to find \( \nu \).

### Final Calculation
After performing the calculations, we find:
\[ T \approx 4.1 \times 10^{-16} \, \text{s} \]
Thus, the frequency is:
\[ \nu \approx 2.44 \times 10^{14} \, \text{s}^{-1} \]

### Conclusion
The frequency of the moving electron in the 3rd orbit of the hydrogen atom is approximately:
\[ \nu \approx 2.44 \times 10^{14} \, \text{Hz} \]