Answer
a) ${\bf 3.5 \times 10^7} \, \text{m/s}$
b) ${\bf 25.54} \;\rm MeV$
Work Step by Step
We have an alpha particle, with charge $ q = 2e $, moving towards a stationary gold nucleus $ ^{197}\text{Au} $ with charge $ q = 79e $.
$$\color{blue}{\bf [a]}$$
In this part, We need to find $v_i$ which is the initial speed of the alpha particle needed to just reach the surface of the gold nucleus before being repelled.
To find the initial speed $ v_i $, we need to use the principle of conservation of energy.
$$
K_i + U_i = K_f + U_f\tag 1
$$
$\bullet $ The alpha particle starts from a large distance, so its initial potential energy $ U_i $ is approximately zero.
$\bullet $ Its initial kinetic energy is $ K_i = \frac{1}{2} m v_i^2 $, where $ m $ is the mass of the alpha particle.
$\bullet $ When the alpha particle reaches the surface of the gold nucleus, it momentarily stops, so its final kinetic energy $ K_f = 0 $.
$\bullet $ Its final potential energy $ U_f $ is due to the Coulomb interaction between the charges of the alpha particle and the gold nucleus $ U_f = \frac{1}{4 \pi \epsilon_0} \frac{(2e)(79e)}{r_f}$.
Substituting into (1)
$$
\frac{1}{2} m v_i^2 + 0 = 0 + \frac{1}{4 \pi \epsilon_0} \frac{(2e)(79e)}{r_f}
$$
Solving for $ v_i $:
$$
v_i^2 = \frac{1}{4 \pi \epsilon_0} \frac{316e^2}{m_\alpha r_f}
$$
$$
v_i = \sqrt{\frac{1}{4 \pi \epsilon_0} \frac{316e^2}{m_\alpha r_f}}
\tag 2$$
Total separation distance $ r_f $ when the alpha particle just touches the gold nucleus:
$$
r_f = r_{\alpha} + r_{\text{Au}} = 1.90 \, \text{fm} + 6.98 \, \text{fm} = 8.88 \, \text{fm}
$$
$$
v_i = \sqrt{ (8.99 \times 10^9 ) \frac{316(1.60 \times 10^{-19} )^2}{ (4 \times 1.67 \times 10^{-27} )(8.88 \times 10^{-15})}}
$$
$$
v_i = \color{red}{\bf 3.5 \times 10^7} \, \text{m/s}
$$
$$\color{blue}{\bf [b]}$$
The initial energy $ K $ of the alpha particle is given by its kinetic energy:
$$
K = \frac{1}{2} m v_i^2
$$
Substitute $ v_i $ from Part (a):
$$
K = \frac{1}{2} (4 \times 1.67\times 10^{-27} ) (3.5 \times 10^7 )^2 =\bf 4.09 \times 10^{-12}\;\rm J
$$
$$K= \color{red}{\bf 25.54} \;\rm MeV$$
