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

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

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

Where:

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

For $ext{^{12}_{6}C}$, the mass number $A = 12$:

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

Calculating the cube root of 12:

$(12)^{1/3} ext{ is approximately } 2.29$

Now we can calculate:

$R ext{ (for } ext{^{12}_{6}C} ext{) } = 1.2 imes 10^{-15} imes 2.29 ext{ m} ext{ (approximately)} \ ext{ } \ R ext{ (for } ext{^{12}_{6}C} ext{) } ext{ is approximately } 2.75 imes 10^{-15} ext{ m. }$

Among the given options, the closest to this value is:

• Option D: $3.12 imes 10^{-15}$ m. Therefore, the radius of a $ext{^{12}_{6}C}$ nucleus is approximately $3.12 imes 10^{-15}$ m.