Let's solve each part step by step, with clear: --- ## (a) **Orbital velocity of the electron** ### **Step 1: Use Centripetal Force = Coulomb Force** The centripetal force for electron's orbit is provided by electrostatic (Coulomb) attraction: \[ \frac{m_e v^2}{a_} = \frac{1}{4\pi\epsilon_} \frac{e^2}{a_^2} \] **Where:** - \( m_e \) = mass of electron = \(9.11 \times 10^{-31}\) kg - \( e \) = charge of electron = \(1.60 \times 10^{-19}\) C - \( a_ \) = Bohr radius = \(5.3 \times 10^{-11}\) m - \( \epsilon_ \) = permittivity of free space = \(8.85 \times 10^{-12}\) F/m --- ### **Step 2: Solve for \( v \)** \[ v = \sqrt{ \frac{1}{4\pi\epsilon_} \frac{e^2}{m_e a_} } \] Plug in values: \[ v = \sqrt{ \frac{1}{4\pi (8.85 \times 10^{-12})} \frac{(1.60 \times 10^{-19})^2}{9.11 \times 10^{-31} \times 5.3 \times 10^{-11}} } \] \[ v \approx 2.19 \times 10^6 \text{ m/s} \] --- **Final Answer (a):** **Orbital velocity of the electron is \(2.19 \times 10^6\) m/s.** --- ## (b) **Ionization energy in eV (is it kinetic, potential, or both?)** ### **Step 1: Find Energy Needed to Remove Electron** Total energy of electron in orbit: \[ E = K + U \] Where, - Kinetic energy, \(K = \frac{1}{2}m_ev^2\) - Potential energy, \(U = -\frac{1}{4\pi\epsilon_} \frac{e^2}{a_}\) From Bohr model: \[ E = -13.6 \text{ eV} \] --- ### **Step 2: Ionization Energy is \(|E|\)** Ionization energy is the energy required to bring the electron from \(E = -13.6\) eV to \(E = \) (free electron). So, ionization energy = \(13.6\) eV. It includes **both kinetic and potential energy**: removing the electron means overcoming the total (not just potential) energy. --- **Final Answer (b):** **Ionization energy is \(13.6\) eV (includes both kinetic and potential energy, but dominated by potential energy).** --- Let me know if you want me to continue with (c), (d), and (e)!

## (c) **Magnetic field at the center of the orbit due to its motion** ### **Step 1: Use Biot-Savart Law** The magnetic field \( B \) at the center of the circular orbit can be calculated using: \[ B = \frac{\mu_0 I}{2r} \] Where: - \( \mu_0 \) = permeability of free space = \( 4\pi \times 10^{-7} \, \text{T m/A} \) - \( I \) = current due to the electron - \( r \) = radius of the orbit ### **Step 2: Find Current \( I \)** Current generated by the electron orbiting is given by: \[ I = \frac{q}{T} \] Where \( T = \frac{2\pi r}{v} \) is the period of the electron's orbit, and \( q \) is the charge of the electron. Substituting for \( T \): \[ T = \frac{2\pi a}{v} \] Thus, \[ I = \frac{e v}{2\pi a} \] Now, plug in \( I \) into the equation for \( B \): \[ B = \frac{\mu_0 \cdot \frac{e v}{2\pi a}}{2a} \] ### **Final Calculation** Substituting \( \mu_0 \), \( e \), \( v \), and \( a \): \[ B = \frac{4\pi \times 10^{-7} \cdot (1.60 \times 10^{-19}) \cdot (2.19 \times 10^6)}{4\pi a^2} \] After calculations, we find: \[ B \approx 9.27 \times 10^{-24} \text{ T} \] --- **Final Answer (c):** **Magnetic field at the center of the orbit is approximately \(9.27 \times 10^{-24}\) T.** --- ## (d) **Magnetic dipole moment** ### **Step 1: Use the Formula for Magnetic Dipole Moment** The magnetic dipole moment \( \mu \) for a current loop is given by: \[ \mu = I \cdot A \] Where \( A = \pi r^2 \) is the area of the circular orbit. ### **Step 2: Calculate the Area and Substitute for \( \mu \)** Substituting for area \( A \) and current \( I \): \[ \mu = \left( \frac{e v}{2\pi a} \right) \cdot \pi a^2 = \frac{e v a}{2} \] Plugging in values for \( e \), \( v \), and \( a \): \[ \mu = \frac{(1.60 \times 10^{-19})(2.19 \times 10^6)(5.3 \times 10^{-11})}{2} \] Calculating gives: \[ \mu \approx 9.27 \times 10^{-24} \, \text{A m}^2 \] --- **Final Answer (d):** **Magnetic dipole moment is approximately \(9.27 \times 10^{-24} \, \text{A m}^2\).** --- ## (e) **Magnetic field strength at a distance of 1 nm from the center on the dipole’s axis** ### **Step 1: Use Magnetic Field Formula for a Dipole** The magnetic field \( B \) at a distance \( r \) along the axis of a magnetic dipole is given by: \[ B = \frac{\mu_0}{2\pi} \frac{2\mu}{r^3} \] ### **Step 2: Substitute Values** Using \( \mu \) from part (d) and substituting \( r = 1 \times 10^{-9} \) m: \[ B = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{2(9.27 \times 10^{-24})}{(1 \times 10^{-9})^3} \] Calculating gives: \[ B \approx 3.71 \times 10^{-7} \text{ T} \] --- **Final Answer (e):** **Magnetic field strength at a distance of 1 nm from the center on the dipole’s axis is approximately \(3.71 \times 10^{-7}\) T.**