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: Determine the radius of the third orbit
According to Bohr's model, the radius of the nth orbit is given by the formula:
\[ r_n = 0.529 \times n^2 \text{ Å} \]
For the third orbit (n = 3) of hydrogen (Z = 1):
\[ r_3 = 0.529 \times 3^2 = 0.529 \times 9 = 4.761 \text{ Å} \]
To convert this to meters:
\[ r_3 = 4.761 \text{ Å} = 4.761 \times 10^{-10} \text{ m} \]

### Step 2: Calculate the circumference of the third orbit
The circumference (C) of the orbit can be calculated using the formula:
\[ C = 2 \pi r \]
Substituting the radius we found:
\[ C = 2 \pi (4.761 \times 10^{-10}) \approx 2.99 \times 10^{-9} \text{ m} \]

### Step 3: Calculate the frequency of revolution
The frequency (f) of the electron can be calculated using the formula:
\[ f = \frac{v}{C} \]
Where \( v \) is the velocity of the electron. Given \( v = 7.2 \times 10^5 \text{ m/s} \):
\[ f = \frac{7.2 \times 10^5}{2.99 \times 10^{-9}} \approx 2.41 \times 10^{14} \text{ revolutions per second (rps)} \]

### Step 4: Determine the number of revolutions made in one second
Since the frequency is already in revolutions per second, the number of revolutions made by the electron in one second is:
\[ \text{Number of revolutions} = f \approx 2.41 \times 10^{14} \]

### Final Answer
The number of revolutions made by the electron around the nucleus in one second is approximately \( 2.41 \times 10^{14} \).

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