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An alpha particle with a kinetic energy of 13.0 MeV makes a head-on collision with a gold nucleus at rest. part A: What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is 79, and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)

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Answer:

8.75 fm

Explanation:

We use this expression

[tex]E=\frac{Ze*(2e)}{4\pi \epsilon_{0} *d}[/tex]

Which is derived from conservation of energy. E is the kinetic energy. d is the distance to find, Ze and 2e are the charges of the gold nucleus and alpha particle respectively. [tex]\frac{1}{4\pi \epsilon_{0} }[/tex] is the coulomb constant which is 9*10^9. Converting the energy to Joules

[tex]13 MeV*\frac{1 J}{6.24*10^{12}MeV}=2.08*10^{-12} J[/tex]

So, solving for distance (having all units in international system, distance ends up in meters)

[tex]d=\frac{9*10^9 *79*2*(1.6*10^{-19})^2}{2.08*10^{-12}} = 8.75*10^{-15} m[/tex]

In fermi:

[tex]8.75*10^{-15} m*\frac{10^{15} fm}{1 m}=8.75 fm[/tex]

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