(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 $$ 180^{\circ} $$ with respect to its original incident direction. (use $$ h c=1240 \mathrm{eV} \cdot \mathrm{nm} $$ ). 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 $$ \ell=1.5 imes 10^{-10} \mathrm{~m} $$, makes a transition from the $$ n=2 $$ state to the $$ n=1 $$ state.

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(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 $$ 180^{\circ} $$ with respect to its original incident direction. (use $$ h c=1240 \mathrm{eV} \cdot \mathrm{nm} $$ ). 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 $$ \ell=1.5 	imes 10^{-10} \mathrm{~m} $$, makes a transition from the $$ n=2 $$ state to the $$ n=1 $$ state.

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### **(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:**
$$
\lambda \approx 24.7 \, \text{nm}
$$

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### **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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