If an alpha particle (two protons and twoneutrons) is given an initial (nonrelativistic) velocityvat a very far distance and is aimed directly at a gold nucleus(Z=79), what is the closest distance d the alpha particle will come to the nucleus? In this problem you can estimate that the mass of the proton m_p is equal to the mass of the neutron and only consider theeffects of a single gold nucleus. Assume that the alpha particle comes close enough so that the nucleus is not substantiallyscreened by inner-shell electrons.
Express your answer in terms ofm_p, the permittivity of free space epsilon_0, the magnitude of the electron charge e, and v.

 r =

Explanation:

To solve this problem we can use conservation of energy,

starting point. With the alpha particle too far from the gold nucleus

          Em₀ = K = ½ m v²

final point. The point of closest approach, whereby the speed of the alpha particle is zero.

          Em_f = U = k q₁q₂ / r

where q₁ is the charge of the alpha particle and q₂ the charge of the Gold nucleus.

Energy is conserved

          Em₀ = Em_f

          ½ m v² = k q₁q₂ / r

          r =

the mass of the particular alpha is

          m_particle = 2 m_proton + 2M_neutron

          m_particle = 4 m_proton

the charge of the alpha and the gold particle are

          q₁ = 2e

          q₂ = 79 e

we substitute

           r =

           r =  

           k =

we substitute

          r =

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y