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		+	### 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.
		+
		+	### 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:
		+	$$K = \frac{(2eBR)^2}{2 \cdot 4m_p} + \frac{(eBR)^2}{2 \cdot 2m_p} = \left(\frac{1}{2} + \frac{1}{4}\right)\frac{(eBR)^2}{m_p} = \frac{3}{4} \frac{(eBR)^2}{m_p}$$
		+
		+	#### Answer
		+
		+	$$K = \frac{3(eBR)^2}{4m_p}$$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### 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.

			### 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:
			$$K = \frac{(2eBR)^2}{2 \cdot 4m_p} + \frac{(eBR)^2}{2 \cdot 2m_p} = \left(\frac{1}{2} + \frac{1}{4}\right)\frac{(eBR)^2}{m_p} = \frac{3}{4} \frac{(eBR)^2}{m_p}$$

			#### Answer

			$$K = \frac{3(eBR)^2}{4m_p}$$

			#### Answer

			[Insert a concise answer or boxed result]