Answer
${\bf 0.928}\;\rm MeV$
Work Step by Step
We need to find the energy of an alpha particle such that its de Broglie wavelength matches the diameter of a $ ^{238}\text{U} $ nucleus.
First, we need to calculate the Radius of the $ ^{238}\text{U} $ Nucleus.
The radius $ r $ of a nucleus with mass number $ A = 238 $ is given by
$$
r = r_0 A^{1/3}
$$
where $ r_0 = 1.2 \, \text{fm} $, and $ A = 238 $:
$$
r = (1.2 \, \text{fm}) \times (238)^{1/3}=\bf 7.436 \;\rm fm
$$
Now we need to determine the Diameter
So, the diameter of the Nucleus $ D $ is twice the radius.
$$
D = 2r = 2 (7.436 \, \text{fm}) =\bf 14.872 \, \rm {fm}
$$
For the de Broglie wavelength $ \lambda $ of the alpha particle to be equal to the diameter of the $ ^{238}\text{U} $ nucleus:
$$
\lambda = D = 14.873, \text{fm}
$$
where
$$
\lambda = \frac{h}{p}
$$
Thus,
$$
p = \frac{h}{\lambda} = \frac{h}{2r}= \frac{h}{D}
$$
where $ h = 6.63 \times 10^{-34} \, \text{Js} $, thus
$$
p = \frac{(6.63 \times 10^{-34}) }{(14.873\times 10^{-15})}
$$
$$
p =\bf 4.458 \times 10^{-20} \, \rm {kg m/s}
$$
Now we need to calculate the Kinetic Energy which is given by
$$ K = \frac{p^2}{2m} $$
Recalling that, the mass of an alpha particle is 4 times the proton's mass.
Thus,
$$
K = \frac{(4.458 \times 10^{-20} )^2}{2 ( 4 \times 1.67 \times 10^{-27} )}
$$
$$
K=\bf 1.48756\times 10^{-13} \, \rm {J}=\color{red}{\bf 0.928}\;\rm MeV
$$
