(e) (I) In a Compton experiment, an X-ray photon of wavelength 0.15 nm collides with a
stationary free electron and scatters backward at an angle of with respect to its
original incident direction. (use ). What is the wavelength of the
scattered photon (in nm )?
(II) Calculate the wavelength of a photon emitted when an electron, trapped in an
infinitely deep square well potential of width , makes a transition from
the state to the state.

Ask by Ortiz Reeves. in Egypt

Jan 05,2025

Question

(e) (I) In a Compton experiment, an X-ray photon of wavelength 0.15 nm collides with a
stationary free electron and scatters backward at an angle of with respect to its
original incident direction. (use ). What is the wavelength of the
scattered photon (in nm )?
(II) Calculate the wavelength of a photon emitted when an electron, trapped in an
infinitely deep square well potential of width , makes a transition from
the state to the state.

Ask by Ortiz Reeves. in Egypt

Jan 05,2025

UpStudy AI · Accepted Answer

Let's solve each part of the problem step-by-step.

---

### **(I) Compton Scattering**

**Given:**
- Initial wavelength of the X-ray photon, $$ \lambda = 0.15 $$ nm
- Scattering angle, $$ \theta = 180^\circ $$
- Compton wavelength of the electron, $$ \lambda_C = \frac{h}{m_e c} \approx 0.02426 $$ nm

**Compton Formula:**
$$
\lambda' = \lambda + \lambda_C (1 - \cos \theta)
$$

**Calculation:**
$$
\cos 180^\circ = -1
$$
$$
\lambda' = 0.15 \, \text{nm} + 0.02426 \, \text{nm} \times (1 - (-1)) = 0.15 \, \text{nm} + 2 \times 0.02426 \, \text{nm} = 0.15 \, \text{nm} + 0.04852 \, \text{nm} = 0.19852 \, \text{nm}
$$

**Final Answer:**
$$
\lambda' \approx 0.199 \, \text{nm}
$$

---

### **(II) Photon Emission from an Electron Transition in an Infinite Square Well**

**Given:**
- Width of the well, $$ \ell = 1.5 \times 10^{-10} $$ m
- Transition from $$ n = 2 $$ to $$ n = 1 $$

**Energy Levels in an Infinite Square Well:**
$$
E_n = \frac{n^2 \pi^2 \hbar^2}{2 m \ell^2}
$$
$$
\Delta E = E_2 - E_1 = \frac{3 \pi^2 \hbar^2}{2 m \ell^2}
$$

**Constants:**
- Planck's constant, $$ h = 6.626 \times 10^{-34} $$ J·s
- Reduced Planck's constant, $$ \hbar = \frac{h}{2\pi} \approx 1.0545718 \times 10^{-34} $$ J·s
- Electron mass, $$ m = 9.10938356 \times 10^{-31} $$ kg
- Speed of light, $$ c = 3 \times 10^8 $$ m/s
- $$ h c = 1240 $$ eV·nm

**Calculation:**

1. **Calculate $$ \Delta E $$:**
   $$
   \Delta E = \frac{3 \pi^2 \hbar^2}{2 m \ell^2} = \frac{3 \pi^2 (1.0545718 \times 10^{-34} \, \text{J·s})^2}{2 \times 9.10938356 \times 10^{-31} \, \text{kg} \times (1.5 \times 10^{-10} \, \text{m})^2}
   $$
   $$
   \Delta E \approx 8.03 \times 10^{-18} \, \text{J}
   $$
   $$
   \Delta E \approx \frac{8.03 \times 10^{-18} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 50.1 \, \text{eV}
   $$

2. **Calculate the Wavelength $$ \lambda $$ of the Emitted Photon:**
   $$
   \lambda = \frac{h c}{\Delta E} = \frac{1240 \, \text{eV·nm}}{50.1 \, \text{eV}} \approx 24.7 \, \text{nm}
   $$

**Final Answer:**
$$
\lambda \approx 24.7 \, \text{nm}
$$

---

### **Summary of Answers:**
1. **Wavelength of the scattered photon:** $$ \boxed{0.199 \, \text{nm}} $$
2. **Wavelength of the emitted photon:** $$ \boxed{24.7 \, \text{nm}} $$
The scattered photon has a wavelength of approximately 0.199 nm, and the emitted photon has a wavelength of about 24.7 nm.

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