The radius of a nucleus of the iron nuclide $^{56}_{26}Fe$ is $4.35 \times 10^{-15}$ m - AQA - A-Level Physics - Question 31 - 2017 - Paper 2

To estimate the radius of the uranium nucleus, we can use the empirical formula for the radius of a nucleus, which is given by:

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

where $R_0$ is a constant approximately equal to $1.2 \times 10^{-15} m$ and $A$ is the mass number of the nucleus.

For the uranium nuclide $^{238}_{92}U$, the mass number $A = 238$:

$R_{U} = R_0 (238)^{1/3}$

Calculating this gives:

$R_{U} = 1.2 \times 10^{-15} m \cdot (238)^{1/3}$

Now, we approximate $(238)^{1/3} \approx 6.2$:

$R_{U} \approx 1.2 \times 10^{-15} m \cdot 6.2 \approx 7.44 \times 10^{-15} m$

From the provided options, the closest value to our calculated radius is:

D. $7.05 \times 10^{-15} m$