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The magnetic flux through a circular orbit of an electron in hydrogen atom is found to be $\Phi = nh/e$ for $n = 1, 2, 3,...$. The magnetic moment of the electron in the $n$th orbit is:
Options
1
$\frac{ne^2h}{4\pi m}$
2
$\frac{neh}{4\pi m}$
3
$\frac{eh}{4\pi m}$
4
$\frac{n^2eh}{4\pi m}$
Correct Answer
$\frac{neh}{4\pi m}$
Solution
1

Electron orbiting with speed $v$, radius $r$: current $I = \frac{ev}{2\pi r}$

$$\mu = IA = \frac{ev}{2\pi r} \cdot \pi r^2 = \frac{evr}{2}$$
2

Bohr quantization: $mvr = \frac{nh}{2\pi}$ $\Rightarrow$ $vr = \frac{nh}{2\pi m}$

$$\mu = \frac{e}{2} \cdot \frac{nh}{2\pi m} = \boxed{\frac{neh}{4\pi m}}$$
$\mu = \frac{evr}{2} = \frac{neh}{4\pi m} = n\mu_B$ (Bohr magneton)
Theory: Atoms and Nuclei
1. Magnetic Moment of Orbiting Electron

An electron revolving in a circular orbit constitutes a current loop. Current $I = e/T = ev/(2\pi r)$ (charge per unit time). Magnetic moment $\mu = IA = \frac{ev}{2\pi r}\cdot\pi r^2 = \frac{evr}{2} = \frac{eL}{2m}$ where $L = mvr$ = angular momentum. So $\mu = \frac{e}{2m}L$. Since $L$ is quantized ($L = n\hbar$): $\mu_n = \frac{e\hbar}{2m}\cdot n = n\mu_B$ where $\mu_B = e\hbar/(2m)$ = Bohr magneton.

2. Bohr Magneton

$\mu_B = \frac{e\hbar}{2m_e} = \frac{eh}{4\pi m_e} = 9.274\times10^{-24}$ J/T. Smallest unit of magnetic moment for an electron. Magnetic moment of electron in $n$th Bohr orbit: $\mu_n = n\mu_B$. Spin magnetic moment of electron: $\mu_s = g_s\mu_B\sqrt{s(s+1)} \approx \mu_B$ (where $g_s \approx 2$, $s = 1/2$). Nuclear magneton: $\mu_N = e\hbar/(2m_p) = 5.05\times10^{-27}$ J/T (proton mass replaces electron mass, so 1836 times smaller).

3. Bohr Model Results

Radius: $r_n = n^2 a_0$ where $a_0 = 0.529$ Å. Energy: $E_n = -13.6/n^2$ eV. Velocity: $v_n = v_1/n$ where $v_1 = 2.18\times10^6$ m/s. Angular momentum: $L_n = n\hbar$. Magnetic moment: $\mu_n = n\mu_B$. For hydrogen-like ions ($Z > 1$): $r_n = n^2 a_0/Z$, $E_n = -13.6Z^2/n^2$ eV, $v_n = Zv_1/n$. Energy of photon emitted: $h\nu = E_i - E_f = 13.6(1/n_f^2 - 1/n_i^2)$ eV... wait: $h\nu = E_i - E_f$ for $n_i > n_f$.

4. Quantum Numbers

Principal quantum number $n$: energy and size. $n = 1, 2, 3...$. Azimuthal $l$: shape of orbital. $l = 0, 1, ..., n-1$. Magnetic $m_l$: orientation. $m_l = -l, ..., 0, ..., +l$. Spin $m_s$: $\pm 1/2$. Total orbitals in shell $n$: $n^2$. Total electrons: $2n^2$. Pauli exclusion: no two electrons have same four quantum numbers. Hunds rule: half-fill before pairing. Aufbau: fill lowest energy first.

5. de Broglie and Bohr

de Broglie: $\lambda = h/p = h/(mv)$. Bohr condition $mvr = n\hbar$ is equivalent to $2\pi r = n\lambda$ (integer wavelengths fit in orbit). For $n$th orbit: $n$ complete waves fit around circumference. This gives a physical meaning to Bohr quantization: standing wave condition for electron wave. Davisson-Germer (1927): confirmed wave nature of electrons by diffraction experiment.

6. Atomic Spectra

Rydberg formula: $\frac{1}{\lambda} = R_H\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ where $R_H = 1.097\times10^7$ m$^{-1}$. Series: Lyman ($n_f=1$, UV), Balmer ($n_f=2$, visible: $H_\alpha=656$ nm red, $H_\beta=486$ nm blue-green), Paschen ($n_f=3$, IR), Brackett ($n_f=4$), Pfund ($n_f=5$). Ionization energy of hydrogen from ground state: 13.6 eV.

7. Radioactivity

$N = N_0 e^{-\lambda t}$. Half-life $t_{1/2} = 0.693/\lambda$. Activity $A = \lambda N$. Alpha decay: $^A_Z X \rightarrow ^{A-4}_{Z-2}Y + ^4_2He$. Beta minus: neutron $\rightarrow$ proton + $e^-$ + $\bar{\nu}_e$. Beta plus: proton $\rightarrow$ neutron + $e^+$ + $\nu_e$. Gamma: no change in $A$ or $Z$ (energy release from excited nucleus). Binding energy: $BE = (Zm_p + Nm_n - M_{nucleus})c^2$. Mass defect $\Delta m = Zm_p + Nm_n - M$.

8. Nuclear Fission and Fusion

Fission: heavy nucleus splits ($^{235}_{92}U + n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3n + \sim200$ MeV). Chain reaction: each fission releases 2–3 neutrons. Critical mass: minimum for sustained chain reaction. Reactor: controlled fission ($k=1$). Bomb: uncontrolled ($k>1$). Fusion: light nuclei combine ($^2H + ^3H \rightarrow ^4He + n + 17.6$ MeV). Requires $\sim10^7$ K. Sun powered by pp chain. Hydrogen bomb: uncontrolled fusion. ITER (fusion reactor): under construction in France.

Frequently Asked Questions
1. What is the relation between angular momentum and magnetic moment?
$\mu = \frac{e}{2m}L$ where $L = mvr$ = orbital angular momentum. This is a classical result: rotating charged particle creates magnetic dipole. The $g$-factor relates magnetic moment to angular momentum: $\mu = g\frac{e}{2m}L$. For orbital motion: $g_L = 1$. For electron spin: $g_s \approx 2$ (from Dirac equation, QED correction gives 2.00232). Measurement of $g$-factor to 10 decimal places: one of most precise tests of quantum electrodynamics.
2. What is the significance of Bohr magneton?
$\mu_B = e\hbar/(2m_e) = 9.274\times10^{-24}$ J/T is the fundamental unit of atomic magnetic moments. Paramagnetism: atoms with unpaired electrons have permanent magnetic moments $\sim \mu_B$. In external field: these align partially → weak attraction. Ferromagnetism: domains of aligned moments (Fe, Ni, Co). Each atom contributes $\sim 2\mu_B$. MRI (Magnetic Resonance Imaging): nuclear magnetic moments (proton $\mu = 2.793\mu_N$) precess in field → detect RF signal → image.
3. How does the flux condition $\Phi = nh/e$ relate to Bohr quantization?
$\Phi = BA = \frac{\mu_0 I}{2r}\cdot\pi r^2 = \frac{\mu_0 ev r}{2}$. Setting $\Phi = nh/e$: $\frac{\mu_0 evr}{2} = \frac{nh}{e}$... this does not directly simplify to Bohr without additional relations. The condition $\Phi = nh/e$ is the magnetic flux quantization condition (similar to Aharonov-Bohm effect and superconductor flux quantization $\Phi = nh/(2e)$ for Cooper pairs). It yields the same quantized orbits as Bohr condition through a different physical argument.
4. What is the magnetic moment of electron in ground state (n=1)?
$\mu_1 = 1 \times \mu_B = \frac{eh}{4\pi m_e} = \frac{1.6\times10^{-19}\times6.63\times10^{-34}}{4\pi\times9.11\times10^{-31}} = \frac{1.06\times10^{-52}}{1.14\times10^{-29}} = 9.27\times10^{-24}$ J/T. This is the Bohr magneton $\mu_B$. For $n=2$: $\mu_2 = 2\mu_B = 1.854\times10^{-23}$ J/T. Magnetic moment increases linearly with orbit number.
5. What happens to magnetic moment when electron jumps to higher orbit?
When electron absorbs photon and jumps from $n_1$ to $n_2$ ($n_2 > n_1$): magnetic moment changes from $\mu_1 = n_1\mu_B$ to $\mu_2 = n_2\mu_B$. The change $\Delta\mu = (n_2-n_1)\mu_B$. For Lyman series ($n_1=1$, $n_2=2,3,...$): $\Delta\mu = \mu_B, 2\mu_B, ...$. For Balmer series ($n_1=2$): smallest change is $\mu_B$ (transition $2\to3$). The increase in magnetic moment is accompanied by increase in orbital angular momentum and decrease in orbital speed (since $v_n \propto 1/n$).
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