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en/10.1.6.md
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| ### Statement | |||
| − | $10.1.6.$ [Insert the problem statement] | ||
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| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
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| 10.1.6. Using a cloud chamber placed in a magnetic field with induction $B$, the elastic scattering of $\alpha$-particles on deuterium nuclei is observed. Find the initial energy of the $\alpha$-particle if the radius of curvature of the initial sections of the trajectories of the nucleus and the $\alpha$-particle after scattering turned out to be equal to $R$. Both trajectories lie in a plane perpendicular to the magnetic field induction. | |||
| ### Solution | |||
| After the collision, the particles move along circular trajectories in a plane perpendicular to the magnetic field $B$ under the action of the Lorentz force. Let us write Newton's second law: | |||
| $$qvB = m\frac{v^2}{r}$$ | |||
| From here, the trajectory radius is: | |||
| $$r = \frac{mv}{qB}$$ | |||
| The magnitude of the particle's momentum $p = mv$ can be expressed from the last formula: | |||
| $$p = qBr$$ | |||
| In an elastic scattering on a stationary deuterium nucleus, both momentum and energy are conserved. Let us write down the masses and charges of the $\alpha$-particle and the deuterium nucleus. An $\alpha$-particle consists of two protons and two neutrons. Since the masses of a proton and a neutron are practically the same, the mass of the $\alpha$-particle is equal to four proton masses, and its charge is equal to double the elementary charge: $m_\alpha = 4m_p$ and $q_\alpha = 2e$. | |||
| The deuterium nucleus consists of one proton and one neutron, so for it we get: $m_D = 2m_p$ and $q_D = e$. | |||
| According to the problem statement, after scattering, both particles have the same trajectory radius $R$ in the magnetic field, so the momentum of the particles will be: | |||
| $$p_\alpha = q_\alpha BR = 2eBR$$ | |||
| $$p_D = q_D BR = eBR$$ | |||
| Let us write the law of conservation of energy: | |||
| $$K = \frac{p_\alpha^2}{2m_\alpha} + \frac{p_D^2}{2m_D}$$ | |||
| Substitute the masses and momenta: | |||
| $$K = \frac{(2eBR)^2}{2 \cdot 4m_p} + \frac{(eBR)^2}{2 \cdot 2m_p} = \left(\frac{1}{2} + \frac{1}{4}\right)\frac{(eBR)^2}{m_p} = \frac{3}{4} \frac{(eBR)^2}{m_p}$$ | |||
| @@ -36,7 +30,3 @@Solution | |||
| #### Answer | |||
| $$K = \frac{3(eBR)^2}{4m_p}$$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $10.1.6.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| 10.1.6. Using a cloud chamber placed in a magnetic field with induction $B$, the elastic scattering of $\alpha$-particles on deuterium nuclei is observed. Find the initial energy of the $\alpha$-particle if the radius of curvature of the initial sections of the trajectories of the nucleus and the $\alpha$-particle after scattering turned out to be equal to $R$. Both trajectories lie in a plane perpendicular to the magnetic field induction. | 10.1.6. Using a cloud chamber placed in a magnetic field with induction $B$, the elastic scattering of $\alpha$-particles on deuterium nuclei is observed. Find the initial energy of the $\alpha$-particle if the radius of curvature of the initial sections of the trajectories of the nucleus and the $\alpha$-particle after scattering turned out to be equal to $R$. Both trajectories lie in a plane perpendicular to the magnetic field induction. | ||
| ### Solution | ### Solution | ||
| After the collision, the particles move along circular trajectories in a plane perpendicular to the magnetic field $B$ under the action of the Lorentz force. Let us write Newton's second law: | After the collision, the particles move along circular trajectories in a plane perpendicular to the magnetic field $B$ under the action of the Lorentz force. Let us write Newton's second law: | ||
| $$qvB = m\frac{v^2}{r}$$ | $$qvB = m\frac{v^2}{r}$$ | ||
| From here, the trajectory radius is: | From here, the trajectory radius is: | ||
| $$r = \frac{mv}{qB}$$ | $$r = \frac{mv}{qB}$$ | ||
| The magnitude of the particle's momentum $p = mv$ can be expressed from the last formula: | The magnitude of the particle's momentum $p = mv$ can be expressed from the last formula: | ||
| $$p = qBr$$ | $$p = qBr$$ | ||
| In an elastic scattering on a stationary deuterium nucleus, both momentum and energy are conserved. Let us write down the masses and charges of the $\alpha$-particle and the deuterium nucleus. An $\alpha$-particle consists of two protons and two neutrons. Since the masses of a proton and a neutron are practically the same, the mass of the $\alpha$-particle is equal to four proton masses, and its charge is equal to double the elementary charge: $m_\alpha = 4m_p$ and $q_\alpha = 2e$. | In an elastic scattering on a stationary deuterium nucleus, both momentum and energy are conserved. Let us write down the masses and charges of the $\alpha$-particle and the deuterium nucleus. An $\alpha$-particle consists of two protons and two neutrons. Since the masses of a proton and a neutron are practically the same, the mass of the $\alpha$-particle is equal to four proton masses, and its charge is equal to double the elementary charge: $m_\alpha = 4m_p$ and $q_\alpha = 2e$. | ||
| The deuterium nucleus consists of one proton and one neutron, so for it we get: $m_D = 2m_p$ and $q_D = e$. | The deuterium nucleus consists of one proton and one neutron, so for it we get: $m_D = 2m_p$ and $q_D = e$. | ||
| According to the problem statement, after scattering, both particles have the same trajectory radius $R$ in the magnetic field, so the momentum of the particles will be: | According to the problem statement, after scattering, both particles have the same trajectory radius $R$ in the magnetic field, so the momentum of the particles will be: | ||
| $$p_\alpha = q_\alpha BR = 2eBR$$ | $$p_\alpha = q_\alpha BR = 2eBR$$ | ||
| $$p_D = q_D BR = eBR$$ | $$p_D = q_D BR = eBR$$ | ||
| Let us write the law of conservation of energy: | Let us write the law of conservation of energy: | ||
| $$K = \frac{p_\alpha^2}{2m_\alpha} + \frac{p_D^2}{2m_D}$$ | $$K = \frac{p_\alpha^2}{2m_\alpha} + \frac{p_D^2}{2m_D}$$ | ||
| Substitute the masses and momenta: | Substitute the masses and momenta: | ||
| $$K = \frac{(2eBR)^2}{2 \cdot 4m_p} + \frac{(eBR)^2}{2 \cdot 2m_p} = \left(\frac{1}{2} + \frac{1}{4}\right)\frac{(eBR)^2}{m_p} = \frac{3}{4} \frac{(eBR)^2}{m_p}$$ | $$K = \frac{(2eBR)^2}{2 \cdot 4m_p} + \frac{(eBR)^2}{2 \cdot 2m_p} = \left(\frac{1}{2} + \frac{1}{4}\right)\frac{(eBR)^2}{m_p} = \frac{3}{4} \frac{(eBR)^2}{m_p}$$ | ||
| @@ -36,7 +30,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $$K = \frac{3(eBR)^2}{4m_p}$$ | $$K = \frac{3(eBR)^2}{4m_p}$$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||