In accordance with the Bohr's model, find the quantum number that characterizes the earth's revolution around the sun in an orbit of radius `1.5xx10^(11)m` with orbital speed `3xx10^4m//s`. (Mass of earth=`6.0xx10^(24)kg`)

Options

A. `5.98xx10^(86)`
B. `2.57xx10^(38)`
C. `8.57xx10^(64)`
D. `2.57xx10^(74)`

Explanation

To find the quantum number that characterizes the Earth's revolution around the Sun using Bohr's model, we can follow these steps:

### Step 1: Write down the given values
- Radius of orbit (r) = \(1.5 \times 10^{11} \, \text{m}\)
- Orbital speed (v) = \(3 \times 10^{4} \, \text{m/s}\)
- Mass of Earth (m) = \(6.0 \times 10^{24} \, \text{kg}\)

### Step 2: Use the Bohr model equation
According to Bohr's model, the angular momentum of an object in circular motion is quantized and given by the formula:
\[ mvr = \frac{nh}{2\pi} \]
where:
- \(n\) = quantum number
- \(h\) = Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))

### Step 3: Rearrange the equation to solve for \(n\)
Rearranging the equation gives:
\[ n = \frac{2\pi mvr}{h} \]

### Step 4: Substitute the known values into the equation
Substituting the values we have:
- \(m = 6.0 \times 10^{24} \, \text{kg}\)
- \(v = 3.0 \times 10^{4} \, \text{m/s}\)
- \(r = 1.5 \times 10^{11} \, \text{m}\)
- \(h = 6.626 \times 10^{-34} \, \text{Js}\)

Calculating \(n\):
\[ n = \frac{2\pi (6.0 \times 10^{24}) (3.0 \times 10^{4}) (1.5 \times 10^{11})}{6.626 \times 10^{-34}} \]

### Step 5: Calculate the numerator
Calculating the numerator:
\[ 2\pi (6.0 \times 10^{24}) (3.0 \times 10^{4}) (1.5 \times 10^{11}) \approx 5.65 \times 10^{41} \]

### Step 6: Calculate \(n\)
Now substituting back into the equation:
\[ n = \frac{5.65 \times 10^{41}}{6.626 \times 10^{-34}} \approx 8.53 \times 10^{74} \]

### Conclusion
The quantum number that characterizes the Earth's revolution around the Sun is:
\[ n \approx 8.53 \times 10^{74} \]