During a famous experiment in 1919, Ernest Rutherford shot doubly ionized helium nuclei (also known as alpha particles) at a gold foil. He discovered that virtually all of the mass of an atom resides in an extremely compact nucleus. Suppose that during such an experiment, an alpha particle far from the foil has a kinetic energy of 6.0 MeV. If the alpha particle is aimed directly at the gold nucleus, and the only force acting on it is the electric force of repulsion exerted on it by the gold nucleus, how close will it approach the gold nucleus before turning back

r = 3.79 10⁻¹⁵ m

Explanation:

For this exercise we can use energy conservation

Starting point. Far from the fixed core

       Em₀ = K

Final point

        = U = k q1 q2 / r

      Em₀ =Em_{f}

       K = k q₁ q₂ / r

       r = k q₂ q₂ / K

Data

     K = 6 Mev = 6 106 eV (1.6 10⁻¹⁹ J / 1 eV) = 9.6 10⁻¹³ J

    Helium

    q₁ = 2 e = 2 1.6 10⁻¹⁹ C = 3.2 10⁻¹⁹ C

     Gold

    q2 = 79 e = 79 1.6 10⁻¹⁹ = 126.4 10⁻¹⁹ C

Let's calculate

      r = 8.99 10⁸ 3.2 10⁻¹⁹ 126.4 10⁻¹⁹ / 9.6 10⁻¹³

      r = 3.79 10⁻¹⁵ m

The answer to the question is;

The alpha particle will approach the gold nucleus up to 3.79 × 10⁻¹⁴ m before turning back.

Explanation:

To solve the question, we note that

There are  79 electrons the gold tom and 2 electrons in the alpha particle which is the helium nuclear

We use the energy conservation principle to obtain

W = = Δ KE

Where

F = Coulomb force

r = Distance of the alpha particle from the gold foil

we get from F = k(Q₁·Q₂)/r²

W = =(-1)×

= -158·k·e² × ⇒ -158·k·e² = = -KE₁

or r = (158·k·e²)/KE₁

Where Δ KE = - KE₁ as KE₂ = 0, particle at rest

Solving gives

r =     = 3.79 × 10⁻¹⁴ m

your answer is c! hope this : )

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