The approximate value of quantum number `n` for the circular orbit of hydrogen of `0.0001 mm` in diameter is

Options

A. 1000
B. 60
C. 10000
D. 31

Explanation

To find the approximate value of the quantum number \( n \) for the circular orbit of hydrogen with a diameter of \( 0.0001 \) mm, we can follow these steps:

### Step 1: Convert the diameter to meters
The diameter given is \( 0.0001 \) mm. We need to convert this to meters:
\[ 0.0001 \text{ mm} = 0.0001 \times 10^{-3} \text{ m} = 10^{-7} \text{ m} \]

### Step 2: Calculate the radius
The radius \( r \) is half of the diameter:
\[ r = \frac{d}{2} = \frac{10^{-7}}{2} = 0.5 \times 10^{-7} \text{ m} \]

### Step 3: Use the formula for the radius of the nth orbit
The radius \( R_n \) of the nth orbit for a hydrogen atom can be expressed as:
\[ R_n = R_0 \cdot n^2 \]
where \( R_0 \) is the ground state radius of the hydrogen atom, approximately \( 0.529 \times 10^{-10} \text{ m} \).

### Step 4: Set up the equation
We know:
\[ R_n = 0.5 \times 10^{-7} \text{ m} \]
Substituting into the equation:
\[ 0.5 \times 10^{-7} = 0.529 \times 10^{-10} \cdot n^2 \]

### Step 5: Solve for \( n^2 \)
Rearranging the equation to solve for \( n^2 \):
\[ n^2 = \frac{0.5 \times 10^{-7}}{0.529 \times 10^{-10}} \]

### Step 6: Calculate \( n^2 \)
Calculating the right-hand side:
\[ n^2 = \frac{0.5 \times 10^{-7}}{0.529 \times 10^{-10}} = \frac{0.5}{0.529} \times 10^{3} \approx 0.946 \times 10^{3} \approx 946 \]

### Step 7: Find \( n \)
Taking the square root to find \( n \):
\[ n \approx \sqrt{946} \approx 30.8 \]
Rounding to the nearest whole number gives:
\[ n \approx 31 \]

### Final Answer
The approximate value of the quantum number \( n \) for the circular orbit of hydrogen with a diameter of \( 0.0001 \) mm is \( n = 31 \).

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