An alpha particle (q=+2eq = +2eq=+2e) is directed head-on at a stationary platinum nucleus (78195Pt^{195}_{78}\text{Pt}78195Pt). The initial kinetic energy of the alpha particle is 7.5 MeV7.5\text{ MeV}7.5 MeV.
Which of the following is the closest estimate for the distance of closest approach d d\,d of the alpha particle to the center of the platinum nucleus?
(Take 14πε0=8.99×109 F m−1\displaystyle \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^9\text{ F m}^{-1}4πε01=8.99×109 F m−1 and the elementary charge e=1.60×10−19 Ce = 1.60 \times 10^{-19}\text{ C}e=1.60×10−19 C.)
1.5×10−14 m1.5 \times 10^{-14}\text{ m}1.5×10−14 m
3.0×10−14 m3.0 \times 10^{-14}\text{ m}3.0×10−14 m
6.0×10−14 m6.0 \times 10^{-14}\text{ m}6.0×10−14 m
3.8×10−16 m3.8 \times 10^{-16}\text{ m}3.8×10−16 m
