Answer
${\bf 4.77 \times 10^7 }\; \text{m/s}$
Work Step by Step
We are asked to calculate the minimum speed of a proton, $ v_1 $, which is sufficient to cause a nuclear excitation that results in the emission of a gamma ray. Specifically, we are modeling the nucleus as a potential well.
We know, from the mentioned graph and the problem statement, that:
$\bullet$ The nucleus consists of a charge $( Z \;e) $, where $ Z = 13 $.
$\bullet$ The proton interacts with the nucleus, and we are considering the proton's motion from infinity down to the nucleus' radius, $ r_2 $, which is approximately $ 4 \; \text{fm} $ (femtometers).
$\bullet$ The proton's initial kinetic energy at infinity equals the energy required to excite the nucleus from $ n = 1 $ (ground state) to $ n = 2 $ (excited state). This energy is $ 7.2 \; \text{MeV} $, which corresponds to the gamma-ray energy emitted.
From the problem statement, the gamma ray wavelength $ \lambda = 1.73 \times 10^{-4} \; \text{nm} $ corresponds to a photon energy:
$$
E_{\text{photon}} = \frac{hc}{\lambda}
$$
This energy is given as:
$$
E_{\text{photon}} = 7.2 \; \text{MeV}
$$
This energy represents the difference in energy between the $ n = 2 $ and $ n = 1 $ states of the nucleus. Therefore, the proton must have at least this much energy when it reaches the nucleus to excite it.
The conservation of energy in this situation indicates that;
$$
K_1+U_1=K_2+U_2
$$
So,
$$
\frac{1}{2} m v_1^2 + 0 = K_2 + \frac{Z e^2}{4 \pi \epsilon_0 r}
$$
Solving for $v_1$;
$$
v_1 = \sqrt{2\left[\dfrac{K_2 + \frac{Z e^2}{4 \pi \epsilon_0 r}}{m}\right]}
$$
Substitute the known:
$$
v_1 = \sqrt{2\left[\dfrac{(7.2\times10^6\times 1.6\times 10^{-19})+ \frac{(9.0 \times 10^9 )(13) (1.6 \times 10^{-19} )^2}{4.0 \times 10^{-15} }}{(1.67\times 10^{-27})}\right]}
$$
$$
v_1 \approx \color{red}{\bf 4.77 \times 10^7 }\; \text{m/s}
$$
This speed ensures that the proton has enough kinetic energy at the nucleus to excite the nucleus from the $ n = 1 $ ground state to the $ n = 2 $ excited state. Hence, the initial speed is
$$v_1\geq {\bf 4.77 \times 10^7 }\; \text{m/s}$$
