Question

It the radius of \({ }^{27}{ }_{13} \mathrm{~A} \ell\) nucleus is $3.6 \mathrm{fm}\( the radius of \){ }^{125}{ }_{52} \mathrm{Te}$ nucleus is nearly equal to (A) \(8 \mathrm{fm}\) (B) \(6 \mathrm{fm}\) (C) \(4 \mathrm{fm}\) (D) \(5 \mathrm{fm}\)

Short answer

The radius of the \({ }^{125}{ }_{52} \mathrm{Te}\) nucleus is approximately \(6 \mathrm{fm}\).

Step-by-step solution

Understand the formula for Nuclear Radius

The formula for nuclear radius is given by: \[R = R_0 A^{1/3}\] where R is the nuclear radius, R_0 is the constant (approximately equal to 1.2 \(\mathrm{fm}\)), and A is the mass number of the nucleus.

Identify the given values

The problem gives us the radius of \({ }^{27}{ }_{13} \mathrm{~A} \ell\) nucleus as \(3.6 \mathrm{fm}\). So, in this case, A = 27 and R = 3.6 \(\mathrm{fm}\).

Calculate the value of R_0

To find the value of R_0, we need to rearrange the nuclear radius formula and substitute the values of A and R: \[R_0 = \frac{R}{A^{1/3}}\] Plug in the given values: \[R_0 = \frac{3.6}{27^{1/3}}\] \[R_0 \approx 1.2 \mathrm{fm}\]

Calculate the radius of \({ }^{125}{ }_{52} \mathrm{Te}\) nucleus

To find the radius of \({ }^{125}{ }_{52} \mathrm{Te}\) nucleus, use the nuclear radius formula. Assign the values A = 125, and R_0 = 1.2 \(\mathrm{fm}\). \[R = R_0 A^{1/3}\] \[R = 1.2 (125)^{1/3}\] \[R \approx 6 \mathrm{fm}\]

Identify the correct option

After calculating the radius of \({ }^{125}{ }_{52} \mathrm{Te}\) nucleus, compare it with the given options. Since the radius is approximately \(6 \mathrm{fm}\), the correct answer is (B) \(6 \mathrm{fm}\).