# Bohr Problem Step-by-Step Let's break the problem step by step and explain everything as as possible. --- ## **Background What is the Bohr Model?** The Bohr Model describes the hydrogen atom as an electron moving in a circular orbit around a proton. The model makes two important assumptions: 1. The centripetal force keeping the electron in a circle is provided by the electrostatic attraction between the electron and proton. 2. The angular momentum of the electron is "quantized" (it can only have specific values). --- ## **Given Equations** From the image, the two main equations are: 1. **Circular Motion (Newton’s Law):** \[ \frac{mv^2}{r} = \frac{Ke^2}{r^2} \] - \(m\) = mass of electron - \(v\) = speed of electron - \(r\) = radius of orbit - \(K\) = Coulomb's constant - \(e\) = charge of electron 2. **Quantized Angular Momentum:** \[ mvr = n \frac{h}{2\pi} \] - \(n\) = quantum number (given as \(n=3\) in the question) - \(h\) = Planck constant --- ## **Physical Constants (from the image)** - \(h = 6.626 \times 10^{-34} \) J·s - \(Ke^2 = 2.307 \times 10^{-28}\) J·m - \(m = 9.11 \times 10^{-31}\) kg --- # **(A) Find the Electron Speed for \( n = 3 \)** ## **Step 1: Find the Radius for \( n = 3 \)** From the angular momentum equation: \[ mvr = n \frac{h}{2\pi} \implies v = \frac{n h}{2\pi m r} \] From the circular motion equation: \[ \frac{mv^2}{r} = \frac{Ke^2}{r^2} \implies mv^2 = \frac{Ke^2}{r} \implies v^2 = \frac{Ke^2}{mr} \] Set the two expressions for \(v\) equal (from above): \[ v = \frac{n h}{2\pi m r} \] Square both sides and substitute the result from the second equation: \[ \left( \frac{n h}{2\pi m r} \right)^2 = \frac{Ke^2}{mr} \] \[ \frac{n^2 h^2}{4\pi^2 m^2 r^2} = \frac{Ke^2}{mr} \] Multiply both sides by \(m r^2\): \[ \frac{n^2 h^2}{4\pi^2 m} = Ke^2 r \] \[ r = \frac{n^2 h^2}{4\pi^2 m Ke^2} \] --- ## **Step 2: Plug in \( n = 3 \) and Constants** \[ r_3 = \frac{3^2 \cdot h^2}{4\pi^2 m Ke^2} \] Now plug in values: - \(n = 3\) - \(h = 6.626 \times 10^{-34}\) J·s - \(m = 9.11 \times 10^{-31}\) kg - \(Ke^2 = 2.307 \times 10^{-28}\) J·m - \(\pi \approx 3.1416\) Calculate numerator: \[ n^2 h^2 = 9 \times (6.626 \times 10^{-34})^2 = 9 \times 4.390 \times 10^{-67} = 3.951 \times 10^{-66} \] Calculate denominator: \[ 4\pi^2 m Ke^2 = 4 \times (3.1416)^2 \times 9.11 \times 10^{-31} \times 2.307 \times 10^{-28} \] \[ = 4 \times 9.8696 \times 9.11 \times 10^{-31} \times 2.307 \times 10^{-28} \] \[ = 39.4784 \times 9.11 \times 10^{-31} \times 2.307 \times 10^{-28} \] \[ = 359.739 \times 10^{-31} \times 2.307 \times 10^{-28} \] \[ = 8.297 \times 10^{-29} \times 2.307 \times 10^{-28} \] \[ = 1.912 \times 10^{-56} \] So, \[ r_3 = \frac{3.951 \times 10^{-66}}{1.912 \times 10^{-56}} \approx 2.067 \times 10^{-10} \text{ m} \] --- ## **Step 3: Find the Speed \(v\) for \(n = 3\)** Recall: \[ v = \frac{n h}{2\pi m r} \] Plug in values: - \(n = 3\) - \(h = 6.626 \times 10^{-34}\) J·s - \(m = 9.11 \times 10^{-31}\) kg - \(r = 2.067 \times 10^{-10}\) m \[ v = \frac{3 \times 6.626 \times 10^{-34}}{2 \times 3.1416 \times 9.11 \times 10^{-31} \times 2.067 \times 10^{-10}} \] Calculate denominator: \[ 2 \times 3.1416 \times 9.11 \times 10^{-31} \times 2.067 \times 10^{-10} = 6.2832 \times 9.11 \times 10^{-31} \times 2.067 \times 10^{-10} = 57.260 \times 10^{-31} \times 2.067 \times 10^{-10} = 1.184 \times 10^{-39} \] Calculate numerator: \[ 3 \times 6.626 \times 10^{-34} = 19.878 \times 10^{-34} \] So, \[ v = \frac{1.9878 \times 10^{-33}}{1.184 \times 10^{-39}} = 1.679 \times 10^{6} \text{ m/s} \] --- ## **Final Answer for (A)** **For quantum number \( n = 3 \), the electron speed \( v \approx 1.68 \times 10^{6} \) m/s.** --- # **(B) Electron Energy for \( n = 3 \) in Electron-Volts (eV)** ## **Step 1: Bohr Energy Formula** The energy at level \( n \) is: \[ E_n = -\frac{m K^2 e^4}{2 \hbar^2 n^2} \] Or, for hydrogen: \[ E_n = -\frac{13.6}{n^2} \text{ eV} \] For \( n = 3 \): \[ E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV} \] --- ## **Final Answer for (B)** **For quantum number \( n = 3 \), the electron energy \( E_3 = -1.51 \) eV.** --- # **Summary Table** | Part | Formula Used | Value for \(n=3\) | |-----------|------------------------------------------------|-----------------------| | (A) Speed | \(v = \frac{n h}{2\pi m r}\) | \(1.68 \times 10^{6}\) m/s | | (B) Energy| \(E_n = -\frac{13.6\text{ eV}}{n^2}\) | \(-1.51\) eV | --- ## **Python Code Example for Calculation** Here's a Python script for these calculations: ```python import numpy as np # Constants h = 6.626e-34 # Planck's constant (J·s) Ke2 = 2.307e-28 # K*e^2 (J·m) m = 9.11e-31 # Mass of electron (kg) n = 3 # Quantum number # Calculate radius numerator = n**2 * h**2 denominator = 4 * np.pi**2 * m * Ke2 r = numerator / denominator # Calculate speed v = n * h / (2 * np.pi * m * r) # Calculate energy in eV E_n = -13.6 / n**2 # eV print("Radius (m):", r) print("Speed (m/s):", v) print("Energy (eV):", E_n) ``` --- ## **Summary** - The Bohr model uses physics and quantum ideas to find the speed and energy of an electron in hydrogen. - For \(n=3\), the speed is about \(1.68 \times 10^6\) m/s and the energy is \(-1.51\) eV. If you have any follow-up questions or want to see more code, let me know!