An electron with `7.2 xx 10^(5) ms^(-1)` velocity rotates in `3^(rd)` orbit of hydrogen atom. Number of revolution made by this electron around nucleus will be :

Options

A. `2.4 xx 10^(10)` r.p.s
B. `2.5 xx 10^(10)` r.p.s
C. `2.4 xx 10^(14)` r.p.s
D. `2.5 xx 10^(15)` r.p.s

Answer

To solve the problem of determining the number of revolutions made by an electron in the third orbit of a hydrogen atom, we can follow these steps:

### Step 1: Identify the given values
- Velocity of the electron, \( v = 7.2 \times 10^5 \, \text{m/s} \)
- Orbit number, \( n = 3 \)

### Step 2: Calculate the radius of the third orbit
The radius of the nth orbit in a hydrogen atom can be calculated using the formula:
\[ r_n = n^2 \times 0.529 \, \text{Å} \]
Where \( 0.529 \, \text{Å} = 0.529 \times 10^{-10} \, \text{m} \).

For \( n = 3 \):
\[ r_3 = 3^2 \times 0.529 \times 10^{-10} \, \text{m} = 9 \times 0.529 \times 10^{-10} \, \text{m} = 4.761 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the time taken for one revolution
The time taken for one complete revolution is given by:
\[ T = \frac{2 \pi r}{v} \]
Substituting the values:
\[ T = \frac{2 \pi \times 4.761 \times 10^{-10} \, \text{m}}{7.2 \times 10^5 \, \text{m/s}} \]

### Step 4: Calculate the number of revolutions per second
The number of revolutions per second \( f \) can be calculated as:
\[ f = \frac{1}{T} = \frac{v}{2 \pi r} \]
Substituting the values:
\[ f = \frac{7.2 \times 10^5 \, \text{m/s}}{2 \pi \times 4.761 \times 10^{-10} \, \text{m}} \]

### Step 5: Perform the calculation
Calculating the denominator:
\[ 2 \pi \times 4.761 \times 10^{-10} \approx 2.99 \times 10^{-9} \, \text{m} \]
Now substituting this back into the equation for \( f \):
\[ f = \frac{7.2 \times 10^5}{2.99 \times 10^{-9}} \approx 2.41 \times 10^{14} \, \text{revolutions per second} \]

### Final Answer
The number of revolutions made by the electron around the nucleus is approximately:
\[ \boxed{2.41 \times 10^{14}} \, \text{revolutions per second} \]