Новое решение

astrosander правка от 1 апр. 2025 г., 01:36

правка #13992 позже →

en/14.5.1.md +45 −0

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		+	### Statement
		+
		+	$14.5.1.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Problem
		+
		+	14.5.1. A stationary particle of mass $ M $ decays into two $ \gamma $-quanta. Determine the mass of each $ \gamma $-quantum.
		+
		+	### Solution
		+
		+	The total energy of the particle before decay:
		+
		+	\[
		+	E = M c^2
		+	\]
		+
		+	After the decay, two identical $ \gamma $-quanta are produced. Since each $ \gamma $-quantum has no rest mass but carries energy, it is convenient to introduce an equivalent relativistic mass $ m $ corresponding to its energy:
		+
		+	\[
		+	E = 2 m c^2
		+	\]
		+
		+	Applying the law of conservation of energy:
		+
		+	\[
		+	M c^2 = 2 m c^2
		+	\]
		+
		+	Dividing both sides by $ c^2 $, we get:
		+
		+	\[
		+	M = 2m \quad \Rightarrow \quad m = \frac{M}{2}
		+	\]
		+
		+	#### Answer
		+
		+	\[
		+	\boxed{m = \frac{M}{2}}
		+	\]
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$14.5.1.$ [Insert the problem statement]

			### Solution

			### Problem

			14.5.1. A stationary particle of mass $ M $ decays into two $ \gamma $-quanta. Determine the mass of each $ \gamma $-quantum.

			### Solution

			The total energy of the particle before decay:

			\[
			E = M c^2
			\]

			After the decay, two identical $ \gamma $-quanta are produced. Since each $ \gamma $-quantum has no rest mass but carries energy, it is convenient to introduce an equivalent relativistic mass $ m $ corresponding to its energy:

			\[
			E = 2 m c^2
			\]

			Applying the law of conservation of energy:

			\[
			M c^2 = 2 m c^2
			\]

			Dividing both sides by $ c^2 $, we get:

			\[
			M = 2m \quad \Rightarrow \quad m = \frac{M}{2}
			\]

			#### Answer

			\[
			\boxed{m = \frac{M}{2}}
			\]

			#### Answer

			[Insert a concise answer or boxed result]