Let `A_(0)` be the area enclined by the orbit in a hydrogen atom .The graph of in `(A_(0) //A_(1))` against `ln(n)`

Options

A. Will not pass through origin
B. Will be a straight line with slope 4
C. will be rectangular hyperbola
D. Will be parabola

Explanation

To solve the problem step by step, we need to analyze the relationship between the areas of the orbits in a hydrogen atom and the principal quantum number \( n \).

### Step 1: Understand the relationship between radius and quantum number
The radius of the nth orbit in a hydrogen atom is given by:
\[ r_n = n^2 r_1 \]
where \( r_1 \) is the radius of the first orbit.

### Step 2: Calculate the area of the orbits
The area \( A_n \) of the nth orbit can be calculated using the formula for the area of a circle:
\[ A_n = \pi r_n^2 \]
Substituting the expression for \( r_n \):
\[ A_n = \pi (n^2 r_1)^2 = \pi n^4 r_1^2 \]

### Step 3: Express the area of the first orbit
The area of the first orbit \( A_1 \) is:
\[ A_1 = \pi r_1^2 \]

### Step 4: Find the ratio of the areas
Now, we can find the ratio of the area of the nth orbit to the area of the first orbit:
\[ \frac{A_n}{A_1} = \frac{\pi n^4 r_1^2}{\pi r_1^2} = n^4 \]

### Step 5: Take the natural logarithm of the ratio
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{A_n}{A_1}\right) = \ln(n^4) = 4 \ln(n) \]

### Step 6: Graph the relationship
The equation \( \ln\left(\frac{A_n}{A_1}\right) = 4 \ln(n) \) indicates that if we plot \( \ln\left(\frac{A_n}{A_1}\right) \) against \( \ln(n) \), we will get a straight line with a slope of 4.

### Conclusion
Thus, the graph of \( \frac{A_0}{A_1} \) against \( \ln(n) \) will be a straight line with a slope of 4.