If R is the radius and A is the mass number, then log R versus log A graph will be

Options

A. a straight linje
B. a parabola
C. an ellipse
D. none of these

Answer

To solve the problem of finding the relationship between log R and log A, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between R and A**:
The radius \( R \) of a nucleus is given by the formula:
\[ R = R_0 A^{1/3} \]
where \( R_0 \) is a constant (approximately \( 1.2 \) Fermi or \( 1.2 \times 10^{-15} \) meters) and \( A \) is the mass number.

2. **Take the logarithm of both sides**:
To find the relationship between \( \log R \) and \( \log A \), we take the logarithm of the equation:
\[ \log R = \log(R_0 A^{1/3}) \]

3. **Apply the properties of logarithms**:
Using the property of logarithms that states \( \log(ab) = \log a + \log b \) and \( \log(a^b) = b \log a \), we can rewrite the equation:
\[ \log R = \log R_0 + \log A^{1/3} = \log R_0 + \frac{1}{3} \log A \]

4. **Rearranging the equation**:
We can rearrange this equation to express \( \log R \) in terms of \( \log A \):
\[ \log R = \log R_0 + \frac{1}{3} \log A \]

5. **Identify the linear relationship**:
This equation can be rearranged to fit the form of a straight line \( y = mx + c \):
\[ \log R = \frac{1}{3} \log A + \log R_0 \]
Here, \( y = \log R \), \( x = \log A \), the slope \( m = \frac{1}{3} \), and the y-intercept \( c = \log R_0 \).

6. **Conclusion about the graph**:
The graph of \( \log R \) versus \( \log A \) will be a straight line with a slope of \( \frac{1}{3} \) and a y-intercept of \( \log R_0 \).

### Final Answer:
The graph of \( \log R \) versus \( \log A \) will be a straight line with a slope of \( \frac{1}{3} \).

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