A fixed target nucleus is approached head-on by various projectiles. The distance of closest approach is $d=k{Z}_1{Z}_2{e}^2/K$. Which statements are correct?

1. AFor an alpha particle of the same energy, $d$ for gold ($Z=79$) is about 6.1 times $d$ for aluminium ($Z=13$)
2. BA proton of $3.0\mathrm{MeV}$ and an alpha particle of $6.0\mathrm{MeV}$ approach the same nucleus to the same distance
3. CA deuteron of $6.0\mathrm{MeV}$ and an alpha particle of $6.0\mathrm{MeV}$ approach the same nucleus to the same distance
4. DDoubling the alpha-particle energy from $6.0$ to $12.0\mathrm{MeV}$ halves $d$