New solution
en/14.5.24.md
+45 −0
| @@ -0,0 +1,45 @@ | |||
| + | ### Statement | ||
| + | |||
| + | $14.5.24.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $14.5.24.$ Prove that a free electron can neither absorb nor emit a photon. | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | To prove this, let us write the laws of conservation of energy and momentum for the "electron-photon" system and show that they are satisfied only for a non-existent photon with frequency $\nu = 0$. | ||
| + | |||
| + | <b>1. Impossibility of absorption:</b> | ||
| + | Let us consider the process in a reference frame where the free electron is initially at rest. | ||
| + | - Before absorption: the energy of the electron is $E_{e1} = mc^2$, its momentum is $p_{e1} = 0$. The energy of the incident photon is $E_{\gamma} = h\nu$, its momentum is $p_{\gamma} = \frac{h\nu}{c}$. | ||
| + | - After absorption: the photon disappears. According to the law of conservation of momentum, the entire momentum is transferred to the electron, so $p_{e2} = \frac{h\nu}{c}$. The total energy of the electron becomes $E_{e2} = \sqrt{p_{e2}^2 c^2 + m^2 c^4}$. | ||
| + | |||
| + | Let us write the law of conservation of energy: | ||
| + | $$E_{e1} + E_{\gamma} = E_{e2}$$ | ||
| + | $$mc^2 + h\nu = \sqrt{\left(\frac{h\nu}{c}\right)^2 c^2 + m^2 c^4}$$ | ||
| + | $$mc^2 + h\nu = \sqrt{(h\nu)^2 + m^2 c^4}$$ | ||
| + | Squaring both sides of the equation: | ||
| + | $$(mc^2)^2 + 2mc^2 h\nu + (h\nu)^2 = (h\nu)^2 + m^2 c^4$$ | ||
| + | After canceling out identical terms, we get: | ||
| + | $$2mc^2 h\nu = 0$$ | ||
| + | Since the electron mass $m \neq 0$, the speed of light $c \neq 0$, and Planck's constant $h \neq 0$, this equality holds true only if **$\nu = 0$**. Photons with zero frequency do not exist; therefore, complete absorption is impossible. | ||
| + | |||
| + | <b>2. Impossibility of emission:</b> | ||
| + | Suppose a free electron at rest emits a photon with frequency $\nu$. | ||
| + | - Before emission: the energy of the system is equal to the rest energy of the electron $E_1 = mc^2$, and the total momentum is zero. | ||
| + | - After emission: a photon appears with momentum $p_{\gamma} = \frac{h\nu}{c}$. To keep the total momentum zero, the electron must acquire a recoil momentum $p_e = -\frac{h\nu}{c}$. | ||
| + | |||
| + | The law of conservation of energy is: | ||
| + | $$mc^2 = \sqrt{\left(-\frac{h\nu}{c}\right)^2 c^2 + m^2 c^4} + h\nu$$ | ||
| + | $$mc^2 = \sqrt{(h\nu)^2 + m^2 c^4} + h\nu$$ | ||
| + | Since $h\nu > 0$, the right side of the equation is always strictly greater than $mc^2$. The equality is possible only if **$\nu = 0$** (meaning no emission occurs). | ||
| + | |||
| + | #### Answer | ||
| + | The simultaneous satisfaction of the laws of conservation of energy and momentum during the absorption or emission of a photon by a free electron is possible only if the photon frequency is $\nu = 0$. Such photons do not exist. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,45 @@ | |||
| ### Statement | |||
| $14.5.24.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $14.5.24.$ Prove that a free electron can neither absorb nor emit a photon. | |||
| ### Solution | |||
| To prove this, let us write the laws of conservation of energy and momentum for the "electron-photon" system and show that they are satisfied only for a non-existent photon with frequency $\nu = 0$. | |||
| <b>1. Impossibility of absorption:</b> | |||
| Let us consider the process in a reference frame where the free electron is initially at rest. | |||
| - Before absorption: the energy of the electron is $E_{e1} = mc^2$, its momentum is $p_{e1} = 0$. The energy of the incident photon is $E_{\gamma} = h\nu$, its momentum is $p_{\gamma} = \frac{h\nu}{c}$. | |||
| - After absorption: the photon disappears. According to the law of conservation of momentum, the entire momentum is transferred to the electron, so $p_{e2} = \frac{h\nu}{c}$. The total energy of the electron becomes $E_{e2} = \sqrt{p_{e2}^2 c^2 + m^2 c^4}$. | |||
| Let us write the law of conservation of energy: | |||
| $$E_{e1} + E_{\gamma} = E_{e2}$$ | |||
| $$mc^2 + h\nu = \sqrt{\left(\frac{h\nu}{c}\right)^2 c^2 + m^2 c^4}$$ | |||
| $$mc^2 + h\nu = \sqrt{(h\nu)^2 + m^2 c^4}$$ | |||
| Squaring both sides of the equation: | |||
| $$(mc^2)^2 + 2mc^2 h\nu + (h\nu)^2 = (h\nu)^2 + m^2 c^4$$ | |||
| After canceling out identical terms, we get: | |||
| $$2mc^2 h\nu = 0$$ | |||
| Since the electron mass $m \neq 0$, the speed of light $c \neq 0$, and Planck's constant $h \neq 0$, this equality holds true only if **$\nu = 0$**. Photons with zero frequency do not exist; therefore, complete absorption is impossible. | |||
| <b>2. Impossibility of emission:</b> | |||
| Suppose a free electron at rest emits a photon with frequency $\nu$. | |||
| - Before emission: the energy of the system is equal to the rest energy of the electron $E_1 = mc^2$, and the total momentum is zero. | |||
| - After emission: a photon appears with momentum $p_{\gamma} = \frac{h\nu}{c}$. To keep the total momentum zero, the electron must acquire a recoil momentum $p_e = -\frac{h\nu}{c}$. | |||
| The law of conservation of energy is: | |||
| $$mc^2 = \sqrt{\left(-\frac{h\nu}{c}\right)^2 c^2 + m^2 c^4} + h\nu$$ | |||
| $$mc^2 = \sqrt{(h\nu)^2 + m^2 c^4} + h\nu$$ | |||
| Since $h\nu > 0$, the right side of the equation is always strictly greater than $mc^2$. The equality is possible only if **$\nu = 0$** (meaning no emission occurs). | |||
| #### Answer | |||
| The simultaneous satisfaction of the laws of conservation of energy and momentum during the absorption or emission of a photon by a free electron is possible only if the photon frequency is $\nu = 0$. Such photons do not exist. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||