Edits to “Problem”, “Answer”
en/14.5.1.md
+1 −11
| @@ -1,12 +1,6 @@ | |||
| − | ### 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 | |||
| @@ -39,7 +33,3 @@Answer | |||
| \[ | |||
| \boxed{m = \frac{M}{2}} | |||
| \] | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,12 +1,6 @@ | |||
| ### Statement | |||
| $14.5.1.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | ### Problem | ||
| $14.5.1.$ A stationary particle of mass \( M \) decays into two \( \gamma \)-quanta. Determine the mass of each \( \gamma \)-quantum. | |||
| ### Solution | ### Solution | ||
| The total energy of the particle before decay: | The total energy of the particle before decay: | ||
| \[ | \[ | ||
| E = M c^2 | E = M c^2 | ||
| \] | \] | ||
| 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: | 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: | ||
| \[ | \[ | ||
| E = 2 m c^2 | E = 2 m c^2 | ||
| \] | \] | ||
| Applying the law of conservation of energy: | Applying the law of conservation of energy: | ||
| \[ | \[ | ||
| M c^2 = 2 m c^2 | M c^2 = 2 m c^2 | ||
| \] | \] | ||
| Dividing both sides by \( c^2 \), we get: | Dividing both sides by \( c^2 \), we get: | ||
| \[ | \[ | ||
| M = 2m \quad \Rightarrow \quad m = \frac{M}{2} | M = 2m \quad \Rightarrow \quad m = \frac{M}{2} | ||
| \] | \] | ||
| #### Answer | #### Answer | ||
| @@ -39,7 +33,3 @@Answer | |||
| \[ | \[ | ||
| \boxed{m = \frac{M}{2}} | \boxed{m = \frac{M}{2}} | ||
| \] | \] | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||