Question

Which of the following figure represents the variation of \(ln (\frac{R}{R_0})\) with \(ln\  A\)?
(if \(R\) = radius of a nucleus and \(A\) = its mass number)

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the radius of a nucleus \( R \) and its mass number \( A \). According to the empirical formula for the radius of a nucleus:

R = R_0 \cdot A^{1/3}

Where \( R_0 \) is a constant approximately equal to 1.2-1.3 fm (femtometers). To investigate the relationship between \( \ln(\frac{R}{R_0}) \) and \( \ln(A) \), we can take the natural logarithm on both sides of the radius equation:

\ln(R) = \ln(R_0 \cdot A^{1/3})

Using the property of logarithms: \ln(xy) = \ln(x) + \ln(y), we can write:

\ln(R) = \ln(R_0) + \ln(A^{1/3})

Again, using the property of logarithms: \ln(x^y) = y \cdot \ln(x), we can express it as:

\ln(R) = \ln(R_0) + \frac{1}{3}\ln(A)

Now, consider the expression \ln(\frac{R}{R_0}):

\ln(\frac{R}{R_0}) = \ln(R) - \ln(R_0)

Substituting the expression we derived:

\ln(\frac{R}{R_0}) = \ln(R_0) + \frac{1}{3}\ln(A) - \ln(R_0)

This simplifies to:

\ln(\frac{R}{R_0}) = \frac{1}{3}\ln(A)

This is a linear relationship where the slope is \(\frac{1}{3}\). Hence, the plot of \ln(\frac{R}{R_0}) vs \ln(A) will be a straight line with a positive slope of \(\frac{1}{3}\).

Therefore, the correct figure is a straight line graph with a positive slope.