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Chapter 17: Problem 48

In \(1911,\) Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of \(6.68 \times 10^{-27} \mathrm{kg},\) a charge of \(+2 e,\) and an initial velocity of \(1.50 \times 10^{7} \mathrm{m} / \mathrm{s}\) is projected head-on toward a gold nucleus with a charge of \(+79 e\), how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)

Short Answer

Expert verified
Answer: The closest distance between the helium nucleus and the gold nucleus during the collision is approximately \(1.14 \times 10^{-14} \mathrm{m}\).

Step by step solution

01

Write down the given information.

The mass of the helium nucleus is \(m_{He} = 6.68 \times 10^{-27} \mathrm{kg}\), its initial velocity is \(v_{0} = 1.50 \times 10^7 \mathrm{m/s}\), and its charge is \(q_{He} = +2e\). The charge of the gold nucleus is \(q_{Au} = +79e\). We can assume the gold nucleus does not move.
02

Calculate the initial kinetic energy of helium nucleus.

We can calculate the initial kinetic energy (\(KE_{0}\)) of the helium nucleus using the following formula: \(KE_0 = \frac{1}{2}m_{He}v_{0}^2\). Since we know the mass of helium and its initial velocity, we can plug these values into the formula: \(KE_0 = \frac{1}{2}(6.68 \times 10^{-27} \mathrm{kg})(1.5 \times 10^7 \mathrm{m/s})^2 \approx 7.54 \times 10^{-13} \mathrm{J}\).
03

Use energy conservation to find the closest distance between the two nuclei.

At the closest point, all Initial kinetic energy of helium nucleus will transform into potential energy due to the interaction between charges of helium and gold nuclei. So, we can write the energy conservation equation as: \(KE_0 = \frac{q_{He}q_{Au}}{4\pi\epsilon_0r}\), where \(r\) represents the distance between the nuclei at the closest point, and \(\epsilon_0\) is the vacuum permittivity (\(8.85\times10^{-12}F/m\)). We know \(KE_0\), \(q_{He}\), \(q_{Au}\), and \(\epsilon_0\), so we can solve for \(r\) using this equation. Rearranging, we get: \(r = \frac{q_{He}q_{Au}}{4\pi\epsilon_0 KE_0}\).
04

Calculate the closest distance.

Now, we can plug the given values into the formula we derived in step 3: \(r = \frac{(2e)(79e)}{4\pi(8.85\times10^{-12}\mathrm{F/m})(7.54 \times 10^{-13}\mathrm{J})}\), \(r \approx 1.14 \times 10^{-14} \mathrm{m}\). The closest distance between the helium nucleus and the gold nucleus is approximately \(1.14 \times 10^{-14} \mathrm{m}\).

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Most popular questions from this chapter

A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the mass of the drop is \(1.0 \times 10^{-13} \mathrm{kg}\) and it remains stationary when the potential difference between the plates is $9.76 \mathrm{kV},$ what is the magnitude of the charge on the drop? (Ignore the small buoyant force on the drop.)
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