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Sagot :
Answer:
r = [tex]\frac{4\pi \epsilon_o}{79} \frac{m_proton \ v^2}{e^2}[/tex]
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 = [tex]\frac{1}{2} \ \frac{m v^2}{k \ q_1q_2}[/tex]
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 = [tex]\frac{1}{2} \frac{4 m_proton \ v^2 }{k \ 2 \ 79 \ e^2}[/tex]
r = [tex]\frac{1}{79} \ \frac{m_proton \ v^2}{ k \ e^2}[/tex]
k = [tex]\frac{1}{4\pi \epsilon_o }[/tex]
we substitute
r = [tex]\frac{4\pi \epsilon_o}{79} \frac{m_proton \ v^2}{e^2}[/tex]
The closest distance the alpha particle will come to the nucleus will be [tex]\frac{4 \pi \epsilon_0 m_pv^2}{79e^2}[/tex].
What is the law of conservation of energy?
According to the law of conservation of energy, energy can not be created nor be destroyed can be transferred from one form to another form.
The given data in the problem is;
q₁ is the charge of the alpha particle =2e
q₂ the charge of the Gold nucleus. = 79 e
We can employ energy conservation to fix this problem.to begin with, the gold nucleus is too far away from the alpha particle. The kinetic energy of the particle;
[tex]\rm KE= \frac{1}{2} mv^2[/tex]
The potential energy of the particle;
[tex]\rm U= \frac{Kq_1q_2}{r}[/tex]
The energy is conserved;
[tex]\rm KE = PE\\\\ \frac{1}{2}mv^2=\frac{ kq_1q_2}{r} \\\\ r= \frac{1}{2}\frac{mv^2 }{q_1q_2}[/tex]
[tex]\rm r= \frac{1}{2}\frac{4m_pv^2}{k \times 2 \times 79 \times e^2} \\\\ \rm r= \frac{1}{79}\frac{m_pv^2}{k \times \times 79 \times e^2} \\\\ k= \frac{1}{4 \pi \epsilon_0} \\\\ \rm r= \frac{4 \pi \epsilon_0 m_pv^2}{79e^2}[/tex]
Hence the closest distance the alpha particle will come to the nucleus will be [tex]\frac{4 \pi \epsilon_0 m_pv^2}{79e^2}[/tex].
To learn more about the law of conservation of energy refer to link;
brainly.com/question/999862
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