The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of `3`:

Options

A. (a)`6.57xx10^(15)`
B. (b)`6.57xx10^(13)`
C. (c)`6.57xx10^(11)`
D. (d)`6.57xx10^(14)`

Answer

To find the number of revolutions per second made by an electron in the first Bohr orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the Bohr Model
In the Bohr model of the hydrogen atom, the electron moves in circular orbits around the nucleus. The centripetal force required for this circular motion is provided by the electrostatic force between the positively charged nucleus and the negatively charged electron.

### Step 2: Use the Formula for Angular Momentum
The angular momentum \( L \) of the electron in the first orbit is given by:
\[ L = mvr = \frac{nh}{2\pi} \]
where:
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron,
- \( r \) is the radius of the orbit,
- \( n \) is the principal quantum number (for the first orbit, \( n = 1 \)),
- \( h \) is Planck's constant.

### Step 3: Calculate the Radius of the First Bohr Orbit
The radius \( r \) of the first Bohr orbit is given by:
\[ r = 0.53 \times 10^{-10} \text{ m} \]

### Step 4: Calculate the Velocity of the Electron
From the angular momentum equation, we can express the velocity \( v \) as:
\[ v = \frac{nh}{2\pi m r} \]
Substituting \( n = 1 \), \( h = 6.63 \times 10^{-34} \text{ J s} \), and \( m = 9.1 \times 10^{-31} \text{ kg} \):
\[ v = \frac{(1)(6.63 \times 10^{-34})}{2\pi(9.1 \times 10^{-31})(0.53 \times 10^{-10})} \]

### Step 5: Calculate the Number of Revolutions per Second
The number of revolutions per second \( f \) is given by:
\[ f = \frac{v}{2\pi r} \]
Substituting the value of \( v \) from the previous step and the radius \( r \):
\[ f = \frac{v}{2\pi(0.53 \times 10^{-10})} \]

### Step 6: Perform the Calculation
After calculating the value of \( v \) and substituting it into the equation for \( f \), we find:
\[ f \approx 6.57 \times 10^{15} \text{ revolutions per second} \]

### Conclusion
Thus, the number of revolutions per second made by the electron in the first Bohr orbit of the hydrogen atom is of the order of \( 10^{15} \).