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

Options

A. will pass through the origin
B. will be a straigth line with slope `4`
C. will be a monotonically increasing nonlinear curve
D. will be a circle

Answer

To solve the problem step by step, we will analyze the area of the orbits in a hydrogen atom according to Bohr's model and how it relates to the logarithmic graph of \( \frac{A_0}{A_1} \) against \( \log(n) \).

### Step 1: Understand the radius of the nth orbit
According to Bohr's model, the radius \( R_n \) of the nth orbit in a hydrogen atom is given by:
\[ R_n = n^2 a_0 \]
where \( a_0 \) is the Bohr radius.

### Step 2: Calculate the area of the nth orbit
The area \( A_n \) of the nth orbit (which is circular) can be calculated using the formula for the area of a circle:
\[ A_n = \pi R_n^2 \]
Substituting the expression for \( R_n \):
\[ A_n = \pi (n^2 a_0)^2 = \pi n^4 a_0^2 \]

### Step 3: Calculate the area of the first orbit
For the first orbit (where \( n = 1 \)):
\[ A_1 = \pi (1^2 a_0)^2 = \pi a_0^2 \]

### Step 4: Formulate the ratio of areas
Now, we can find the ratio of the area of the nth orbit to the area of the first orbit:
\[ \frac{A_n}{A_1} = \frac{\pi n^4 a_0^2}{\pi a_0^2} = n^4 \]

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

### Step 6: Analyze the relationship
This shows that \( \log\left(\frac{A_n}{A_1}\right) \) is directly proportional to \( \log(n) \). Thus, we can express this relationship as:
\[ \log\left(\frac{A_n}{A_1}\right) = 4 \log(n) \]

### Step 7: Graphical representation
When we plot \( \log\left(\frac{A_n}{A_1}\right) \) on the y-axis against \( \log(n) \) on the x-axis, we will get a straight line passing through the origin with a slope of 4.

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