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Author: quantagate

_posts/2025-03-09-angular-momentum.md

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+---
+title: 'Angular Momentum'
+date: 2025-03-09
+permalink: /posts/2025/09/angular-momentum/
+tags:
+  - quantum
+  - math
+  - physics
+---
+
+This is one of the key postulates of **Bohr’s atomic model**, stating that the **angular momentum** $L$ of an electron in a stable orbit is **quantized**, i.e., it can only take specific discrete values depending on an integer $n$.
+
+$$
+L = n \cdot \frac{h}{2\pi}
+$$
+
+---
+
+### 🧠 Goal of the Explanation:
+
+We’ll explain:
+
+1. **What this equation represents** (in terms of physics)
+2. **Why it’s important** (Bohr’s postulate)
+3. **How it came to be** (historical + mathematical reasoning)
+4. **What math/physics concepts it uses**
+
+---
+
+## 🧪 1. What does the equation mean?
+
+* $L$ is the **angular momentum** of an electron orbiting the nucleus.
+* $h$ is **Planck’s constant**:
+
+  $$
+  h \approx 6.626 \times 10^{-34} \, \text{Js}
+  $$
+* $2\pi$ appears because we’re dealing with **circular motion**.
+* $\frac{h}{2\pi}$ is often written as $\hbar$ (pronounced "h-bar"):
+
+  $$
+  \hbar = \frac{h}{2\pi}
+  $$
+* $n \in \mathbb{Z^+}$ is the **principal quantum number**: 1, 2, 3, …
+
+Thus, Bohr said:
+
+> “The angular momentum of an electron in orbit is quantized and must be an integer multiple of $\hbar$.”
+
+---
+
+## 🔁 2. Why quantize angular momentum?
+
+Bohr’s atomic model was built to explain **why electrons don’t spiral into the nucleus** and why **hydrogen has a discrete line spectrum**.
+
+In classical physics:
+
+* An electron in orbit should emit radiation continuously.
+* It should lose energy and spiral inward — but atoms are stable!
+* So Bohr postulated: **Electrons can only exist in orbits where their angular momentum is quantized**.
+
+That’s where the equation:
+
+$$
+L = n \cdot \frac{h}{2\pi}
+$$
+
+comes in — it’s a **quantization condition**.
+
+---
+
+## ⚙️ 3. How did Bohr come up with it?
+
+Bohr didn’t derive it purely mathematically — it was a **bold postulate** inspired by the new ideas of **quantum theory**, particularly **Planck's** and **Einstein’s** work on **quantized energy**.
+
+But later it was justified using **de Broglie’s hypothesis** and **wave mechanics**.
+
+---
+
+## 🔁 4. Deeper Mathematical Justification (via de Broglie)
+
+After Bohr, in 1924, **Louis de Broglie** suggested that **electrons have wave properties**, and their **orbits must fit whole wavelengths**:
+
+$$
+\text{Circumference of orbit} = n \lambda
+\Rightarrow 2\pi r = n \lambda
+$$
+
+Since de Broglie said:
+
+$$
+\lambda = \frac{h}{p} = \frac{h}{mv}
+$$
+
+Substitute:
+
+$$
+2\pi r = n \cdot \frac{h}{mv}
+\Rightarrow mvr = \frac{nh}{2\pi}
+\Rightarrow L = mvr = n \cdot \frac{h}{2\pi}
+$$
+
+🧠 **So from de Broglie’s wave-particle duality**, we get Bohr’s angular momentum quantization as a **natural result of standing waves**!
+
+---
+
+## 📐 5. What is Angular Momentum? (Quick Refresher)
+
+In circular motion:
+
+$$
+L = m v r
+$$
+
+Where:
+
+* $m$ = mass of electron
+* $v$ = orbital speed
+* $r$ = radius of orbit
+
+Bohr said this quantity $L$ can **only** be $n \cdot \hbar$, not just anything.
+
+---
+
+## 🔢 Summary:
+
+| Concept           | Meaning                                                                                         |
+| ----------------- | ----------------------------------------------------------------------------------------------- |
+| **Equation**      | $L = n \cdot \frac{h}{2\pi}$                                                                    |
+| **Describes**     | Quantized angular momentum of electrons in Bohr orbits                                          |
+| **Postulated by** | Niels Bohr (1913)                                                                               |
+| **Supported by**  | de Broglie’s wave hypothesis (1924)                                                             |
+| **Key idea**      | Electron orbit must accommodate a whole number of de Broglie wavelengths — standing waves only! |
+
+---
+
+## step-by-step derivations 
+
+1. **Radius of Bohr orbit $r_n$**
+2. **Energy of electron in $n^\text{th}$ orbit $E_n$**
+3. **(Bonus)**: Visual of the **standing wave condition** (why angular momentum is quantized)
+
+---
+
+## 🔷 1. Derivation of Radius $r_n$
+
+We’ll derive the **Bohr radius** formula starting from physics principles:
+
+### 🔹 Step 1: Use centripetal force = electrostatic force
+
+An electron moves in a circular orbit, so:
+
+$$
+\text{Centripetal force} = \text{Coulomb (electrostatic) force}
+$$
+
+$$
+\frac{mv^2}{r} = \frac{k e^2}{r^2}
+$$
+
+Where:
+
+* $m$ = electron mass
+* $v$ = velocity
+* $r$ = orbit radius
+* $e$ = electron charge
+* $k = \frac{1}{4\pi\varepsilon_0}$ (Coulomb constant)
+
+---
+
+### 🔹 Step 2: Use Bohr’s angular momentum quantization
+
+$$
+L = mvr = n \cdot \frac{h}{2\pi}
+\Rightarrow v = \frac{n h}{2\pi m r}
+$$
+
+---
+
+### 🔹 Step 3: Substitute velocity $v$ into force equation
+
+$$
+\frac{m}{r} \left(\frac{n h}{2\pi m r}\right)^2 = \frac{k e^2}{r^2}
+$$
+
+Simplify:
+
+$$
+\frac{n^2 h^2}{4\pi^2 m r^3} = \frac{k e^2}{r^2}
+$$
+
+Multiply both sides by $r^3$:
+
+$$
+\frac{n^2 h^2}{4\pi^2 m} = k e^2 r
+$$
+
+Solve for $r$:
+
+$$
+r = \frac{n^2 h^2}{4\pi^2 m k e^2}
+$$
+
+This is the **Bohr radius formula**. For hydrogen ($Z = 1$):
+
+$$
+r_n = n^2 \cdot a_0
+$$
+
+Where $a_0 = \frac{h^2}{4\pi^2 m k e^2} \approx 0.529 \times 10^{-10} \, \text{m}$
+
+---
+
+## 🔷 2. Derivation of Energy $E_n$
+
+Total energy of the electron is:
+
+$$
+E = \text{Kinetic Energy} + \text{Potential Energy}
+$$
+
+### 🔹 Step 1: Kinetic energy
+
+From above:
+
+$$
+\frac{mv^2}{r} = \frac{ke^2}{r^2} \Rightarrow mv^2 = \frac{ke^2}{r}
+$$
+
+So:
+
+$$
+K.E. = \frac{1}{2} mv^2 = \frac{1}{2} \cdot \frac{ke^2}{r}
+$$
+
+---
+
+### 🔹 Step 2: Potential energy
+
+Potential energy in a Coulomb field:
+
+$$
+U = - \frac{ke^2}{r}
+$$
+
+---
+
+### 🔹 Step 3: Total energy
+
+$$
+E = K.E. + U = \frac{1}{2} \cdot \frac{ke^2}{r} - \frac{ke^2}{r} = -\frac{ke^2}{2r}
+$$
+
+Now substitute $r$ from radius formula:
+
+$$
+r = \frac{n^2 h^2}{4\pi^2 m k e^2}
+\Rightarrow \frac{1}{r} = \frac{4\pi^2 m k e^2}{n^2 h^2}
+$$
+
+So:
+
+$$
+E_n = -\frac{1}{2} ke^2 \cdot \frac{4\pi^2 m k e^2}{n^2 h^2}
+= -\frac{2\pi^2 m k^2 e^4}{n^2 h^2}
+$$
+
+This gives the **energy levels** of the hydrogen atom:
+
+$$
+E_n = -\frac{13.6\, \text{eV}}{n^2}
+$$
+
+---
+
+## 🌀 3. Diagram: Standing Wave Condition
+
+Let’s describe how **standing wave = quantized orbits**:
+
+### ⚛ De Broglie Hypothesis:
+
+Electron has **wavelength**:
+
+$$
+\lambda = \frac{h}{mv}
+$$
+
+For a **stable orbit**, the circumference must fit a whole number of wavelengths:
+
+$$
+2\pi r = n \lambda
+$$
+
+Otherwise, the wave interferes destructively and **cancels itself**.
+
+This directly leads to:
+
+$$
+mvr = n \cdot \frac{h}{2\pi}
+\Rightarrow L = n\hbar
+$$
+
+✅ This matches Bohr’s postulate!
+
+---
+
+### 📊 Visualization (text-based)
+
+| Orbit Number | Wave Fit Around Orbit                     |
+| ------------ | ----------------------------------------- |
+| $n = 1$      | One full wave around the circle — 🔁🌊    |
+| $n = 2$      | Two full waves around the circle — 🔁🌊🌊 |
+| $n = 3$      | Three full waves — 🔁🌊🌊🌊               |
+
+In each case, the wave **constructively interferes**, making a **stable standing wave**.
+
+---
+
+## ✅ Summary
+
+| Concept                    | Formula                                          |
+| -------------------------- | ------------------------------------------------ |
+| Radius of orbit            | $r_n = \frac{n^2 h^2}{4\pi^2 m k e^2} = n^2 a_0$ |
+| Bohr radius                | $a_0 = 0.529 \times 10^{-10} \, \text{m}$        |
+| Angular momentum quantized | $L = n \cdot \frac{h}{2\pi}$                     |
+| Energy of orbit            | $E_n = -\frac{13.6}{n^2} \, \text{eV}$           |
+
+---