Новое решение

Alexphysics правка от 23 июн. 2026 г., 20:46

правка #19127 позже →

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		+	### Statement
		+
		+	$14.4.16.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Data:
		+
		+	Proton: kinetic energy$ E = 10^6\ \text{eV} = 1\ \text{MeV}$, deflection angle $\alpha_p = 0.05\ \text{rad}$
		+	Electron: same kinetic energy$ E = 1\ \text{MeV}$
		+	Rest masses: $m_e c^2 = 0.511\ \text{MeV}, m_p c^2 = 938\ \text{MeV}$
		+
		+	Relationship between deflection and momentum
		+
		+	In a plane capacitor with a uniform transverse electric field, a particle of charge e receives a transverse impulse $\Delta p_\perp = eE_\text{field}\,\Delta t$ The flight time is $\Delta t \approx L / v $
		+	where L is the length of the plates and v is the initial velocity. For small angles, the deflection is:
		+
		+	$\alpha \approx \frac{\Delta p_\perp}{p} \propto \frac{1}{p v}$
		+
		+	where p is the initial momentum and v is the velocity. The same geometry and field imply$ \alpha \propto 1/(p v)$
		+
		+	Calculation of p v for a particle with kinetic energy E
		+
		+	In relativity, the total energy is $E_{\text{total}} = E + mc^2$ The momentum satisfies$ (pc)^2 = E_{\text{total}}^2 - (mc^2)^2$ The velocity is$ v/c = pc / E_{\text{total}} $
		+
		+	Multiplying:
		+
		+	$p v = \frac{p^2 c^2}{E_{\text{total}}} = \frac{(E + mc^2)^2 - (mc^2)^2}{E + mc^2} = \frac{E(E + 2mc^2)}{E + mc^2}$
		+
		+	Thus, the product pv is expressed in terms of E and the mass m:
		+
		+	$p v = \frac{E(E + 2mc^2)}{E + mc^2}$
		+
		+	Comparison of the angles
		+
		+	Since$ \alpha \propto 1/(p v)$ the ratio between the deflection angle of the electron and that of the proton is the inverse of the ratio of their p v products:
		+
		+	$\frac{\alpha_e}{\alpha_p} = \frac{(p v)_p}{(p v)_e}$
		+
		+	Substituting the expressions:
		+
		+	$\frac{\alpha_e}{\alpha_p} = \frac{ \dfrac{E(E + 2m_pc^2)}{E + m_pc^2} }{ \dfrac{E(E + 2m_ec^2)}{E + m_ec^2} }
		+	= \frac{E + m_ec^2}{E + 2m_ec^2} \cdot \frac{E + 2m_pc^2}{E + m_pc^2}$
		+
		+	and for small angles $(\alpha_p \ll 1, \alpha_e \ll 1)$ we can approximate$ \tan \alpha \approx \alpha$
		+
		+	Numerical calculation
		+
		+	With$ E = 1\ \text{MeV}, m_ec^2 = 0.511\ \text{MeV}, m_pc^2 = 938\ \text{MeV}$
		+
		+	$\frac{m_ec^2 + E}{2m_ec^2 + E} = \frac{0.511 + 1}{1.022 + 1} = \frac{1.511}{2.022} \approx 0.7473$
		+
		+	$\frac{2m_pc^2 + E}{m_pc^2 + E} = \frac{1876 + 1}{938 + 1} = \frac{1877}{939} \approx 1.9989 \approx 2.00$.
		+
		+	The product is$ 0.7473 \times 1.9989 \approx 1.494$.
		+
		+	For $\alpha_p = 0.05\ \text{rad}$:
		+
		+	$\alpha_e \approx 1.494 \times 0.05 \approx 0.0747\ \text{rad} \approx 0.075\ \text{rad}$.
		+
		+	Result
		+
		+	$\boxed{\alpha_e \approx 0.075\ \text{rad}}$.
		+
		+	The electron deflects approximately 50% more than the proton because, at the same kinetic energy, its smaller mass makes the product p v smaller, thereby increasing the deflection.
		+
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$14.4.16.$ [Insert the problem statement]

			### Solution

			Data:

			Proton: kinetic energy$ E = 10^6\ \text{eV} = 1\ \text{MeV}$, deflection angle $\alpha_p = 0.05\ \text{rad}$
			Electron: same kinetic energy$ E = 1\ \text{MeV}$
			Rest masses: $m_e c^2 = 0.511\ \text{MeV}, m_p c^2 = 938\ \text{MeV}$

			Relationship between deflection and momentum

			In a plane capacitor with a uniform transverse electric field, a particle of charge e receives a transverse impulse $\Delta p_\perp = eE_\text{field}\,\Delta t$ The flight time is $\Delta t \approx L / v $
			where L is the length of the plates and v is the initial velocity. For small angles, the deflection is:

			$\alpha \approx \frac{\Delta p_\perp}{p} \propto \frac{1}{p v}$

			where p is the initial momentum and v is the velocity. The same geometry and field imply$ \alpha \propto 1/(p v)$

			Calculation of p v for a particle with kinetic energy E

			In relativity, the total energy is $E_{\text{total}} = E + mc^2$ The momentum satisfies$ (pc)^2 = E_{\text{total}}^2 - (mc^2)^2$ The velocity is$ v/c = pc / E_{\text{total}} $

			Multiplying:

			$p v = \frac{p^2 c^2}{E_{\text{total}}} = \frac{(E + mc^2)^2 - (mc^2)^2}{E + mc^2} = \frac{E(E + 2mc^2)}{E + mc^2}$

			Thus, the product pv is expressed in terms of E and the mass m:

			$p v = \frac{E(E + 2mc^2)}{E + mc^2}$

			Comparison of the angles

			Since$ \alpha \propto 1/(p v)$ the ratio between the deflection angle of the electron and that of the proton is the inverse of the ratio of their p v products:

			$\frac{\alpha_e}{\alpha_p} = \frac{(p v)_p}{(p v)_e}$

			Substituting the expressions:

			$\frac{\alpha_e}{\alpha_p} = \frac{ \dfrac{E(E + 2m_pc^2)}{E + m_pc^2} }{ \dfrac{E(E + 2m_ec^2)}{E + m_ec^2} }
			= \frac{E + m_ec^2}{E + 2m_ec^2} \cdot \frac{E + 2m_pc^2}{E + m_pc^2}$

			and for small angles $(\alpha_p \ll 1, \alpha_e \ll 1)$ we can approximate$ \tan \alpha \approx \alpha$

			Numerical calculation

			With$ E = 1\ \text{MeV}, m_ec^2 = 0.511\ \text{MeV}, m_pc^2 = 938\ \text{MeV}$

			$\frac{m_ec^2 + E}{2m_ec^2 + E} = \frac{0.511 + 1}{1.022 + 1} = \frac{1.511}{2.022} \approx 0.7473$

			$\frac{2m_pc^2 + E}{m_pc^2 + E} = \frac{1876 + 1}{938 + 1} = \frac{1877}{939} \approx 1.9989 \approx 2.00$.

			The product is$ 0.7473 \times 1.9989 \approx 1.494$.

			For $\alpha_p = 0.05\ \text{rad}$:

			$\alpha_e \approx 1.494 \times 0.05 \approx 0.0747\ \text{rad} \approx 0.075\ \text{rad}$.

			Result

			$\boxed{\alpha_e \approx 0.075\ \text{rad}}$.

			The electron deflects approximately 50% more than the proton because, at the same kinetic energy, its smaller mass makes the product p v smaller, thereby increasing the deflection.


			#### Answer

			[Insert a concise answer or boxed result]