The number of revolutions done by an electron 'e' in one second in the first orbit of hydrogen atom is

Options

A. `6.57 xx 10^(15)`
B. `6.57 xx 10^(13)`
C. `1000`
D. `6.57 xx 10^(14)`.

Explanation

To find the number of revolutions done by an electron in one second in the first orbit of the hydrogen atom, we can follow these steps:

### Step 1: Understand Bohr's Model
In Bohr's model of the hydrogen atom, the angular momentum of the electron in the nth orbit is given by the formula:
\[ M v r = \frac{n h}{2 \pi} \]
where:
- \(M\) = mass of the electron
- \(v\) = velocity of the electron
- \(r\) = radius of the orbit
- \(n\) = principal quantum number (for the first orbit, \(n = 1\))
- \(h\) = Planck's constant

### Step 2: Set Up the Equation for the First Orbit
For the first orbit (\(n = 1\)):
\[ M v r = \frac{1 \cdot h}{2 \pi} \]

### Step 3: Relate Linear Velocity to Angular Velocity
The linear velocity \(v\) can be expressed in terms of angular velocity \(\omega\):
\[ v = r \omega \]
Substituting this into the angular momentum equation gives:
\[ M r^2 \omega = \frac{h}{2 \pi} \]

### Step 4: Express Angular Velocity in Terms of Frequency
The angular velocity \(\omega\) is related to the frequency \( \nu \) (number of revolutions per second) by:
\[ \omega = 2 \pi \nu \]
Substituting this into the equation:
\[ M r^2 (2 \pi \nu) = \frac{h}{2 \pi} \]

### Step 5: Solve for Frequency \(\nu\)
Rearranging the equation to solve for \(\nu\):
\[ \nu = \frac{h}{4 \pi^2 M r^2} \]

### Step 6: Substitute Known Values
Now, we need to substitute the known values:
- \(h = 6.63 \times 10^{-34} \, \text{Js}\)
- \(M = 9.1 \times 10^{-31} \, \text{kg}\)
- The radius of the first orbit \(r = 0.53 \, \text{Å} = 0.53 \times 10^{-10} \, \text{m}\)

Calculating \(r^2\):
\[ r^2 = (0.53 \times 10^{-10})^2 = 2.809 \times 10^{-21} \, \text{m}^2 \]

### Step 7: Calculate \(\nu\)
Substituting the values into the frequency equation:
\[ \nu = \frac{6.63 \times 10^{-34}}{4 \pi^2 (9.1 \times 10^{-31}) (2.809 \times 10^{-21})} \]

Calculating the denominator:
\[ 4 \pi^2 \approx 39.478 \]
\[ 4 \pi^2 \times 9.1 \times 10^{-31} \times 2.809 \times 10^{-21} \approx 1.086 \times 10^{-50} \]

Now calculating \(\nu\):
\[ \nu \approx \frac{6.63 \times 10^{-34}}{1.086 \times 10^{-50}} \approx 6.1 \times 10^{15} \, \text{revolutions per second} \]

### Final Answer
The number of revolutions done by an electron in one second in the first orbit of the hydrogen atom is approximately:
\[ \nu \approx 6.57 \times 10^{15} \, \text{revolutions per second} \]