The radius of a uranium $^{238}_{92}U$ nucleus is $7.75 \times 10^{-15}$ m - AQA - A-Level Physics - Question 30 - 2018 - Paper 2

To determine the radius of the $^{12}_{6}C$ nucleus, we can use the formula for the radius of a nucleus:

$R = R_0 A^{1/3}$

where:

• $R$ is the radius of the nucleus,
• $R_0$ is a proportionality constant (approximately $1.2 \times 10^{-15}$ m),
• $A$ is the mass number of the nucleus.

For the carbon nucleus $^{12}_{6}C$, the mass number $A = 12$. Thus, we calculate:

$R = 1.2 \times 10^{-15} \times (12)^{1/3}$

Calculating $(12)^{1/3}$ gives approximately $2.29$. This leads to:

$R \approx 1.2 \times 10^{-15} \times 2.29 \approx 2.75 \times 10^{-15} \text{ m}$

This value suggests that the closest option is D: $3.12 \times 10^{-15}$ m.