Let `A_(n)` be the area enclosed by the `n^(th)` orbit in a hydrogen atom. The graph of `l n (A_(n)//A_(t))` against In `(n)`

Options

A. will pass through origin
B. will be a straight line with slope 4
C. will be a parabola
D. will be a circle

Answer

To solve the problem, we need to analyze the area enclosed by the nth orbit in a hydrogen atom and how it relates to the quantum number \( n \).

### Step 1: Understanding the Radius of the nth Orbit
The radius of the nth orbit in a hydrogen atom is given by the formula:
\[ r_n = r_0 n^2 \]
where \( r_0 \) is a constant.

### Step 2: Calculating the Area of the nth Orbit
The area \( A_n \) enclosed by 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 \( r_n \):
\[ A_n = \pi (r_0 n^2)^2 = \pi r_0^2 n^4 \]

### Step 3: Expressing the Area in Terms of a Reference Area
Let’s denote the area of the first orbit as \( A_1 \):
\[ A_1 = \pi r_0^2 (1^4) = \pi r_0^2 \]
Now, we can express the area \( A_n \) in terms of \( A_1 \):
\[ A_n = A_1 n^4 \]

### Step 4: Taking the Ratio of Areas
Now, we can take the ratio of the area of the nth orbit to the area of the first orbit:
\[ \frac{A_n}{A_1} = n^4 \]

### Step 5: Taking the Natural Logarithm
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{A_n}{A_1}\right) = \ln(n^4) \]
Using the properties of logarithms, we can simplify this to:
\[ \ln\left(\frac{A_n}{A_1}\right) = 4 \ln(n) \]

### Step 6: Rearranging the Equation
Rearranging the equation, we can express it as:
\[ \ln(A_n) - \ln(A_1) = 4 \ln(n) \]
This indicates that if we plot \( \ln(A_n) \) against \( \ln(n) \), we will get a straight line with a slope of 4.

### Step 7: Analyzing the Graph
The graph of \( \ln\left(\frac{A_n}{A_1}\right) \) against \( \ln(n) \) will be a straight line passing through the origin with a slope of 4.

### Conclusion
Thus, the graph of \( \ln\left(\frac{A_n}{A_1}\right) \) against \( \ln(n) \) is a straight line with a slope of 4.