In the Bohr model of the hydrogen atom, an electron revolves in the first orbit of radius $5.29\times10^{-11}\ \text{m}$ with a speed of $2.19\times10^6\ \text{m/s}$. Calculate the orbital magnetic moment of the electron in this orbit, and compare it with the value of the Bohr magneton, $9.27\times10^{-24}\ \text{A}\cdot\text{m}^2$.

In the Bohr model of the hydrogen atom, an electron revolves in the first orbit of radius 5.29×10⁻¹¹m with a s

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5.29\times10^{-11}\ \text{m}

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Concept understanding — Bohr Magneton

The first time you meet the Bohr magneton, it can feel like a random number with a weird unit. But it is actually one of the most natural constants in atomic physics — it is the smallest possible magnetic moment that an electron can have, simply by moving around.

The intuition: a tiny current loop

Imagine an electron whizzing around a nucleus in a circular orbit. A moving charge is an electric current. A loop of current is a tiny magnet — it has a magnetic moment. The magnetic moment tells you how strong that little magnet is and how it will twist in an external magnetic field.

For a single electron in a circular orbit, the magnetic moment turns out to be proportional to the orbital angular momentum. The constant of proportionality is the Bohr magneton. In other words, if you know how much angular momentum the electron has, you multiply by $\mu_B$ to get its magnetic moment.

The same idea applies to the electron's spin. Even though spin is not a literal spinning motion, it also produces a magnetic moment, and the natural unit for that is again $\mu_B$ (though the exact factor is slightly different — the spin g-factor is about 2).

So the Bohr magneton is the fundamental quantum of magnetic moment for an electron. Any atomic magnetic moment is an integer multiple (or a simple fraction) of $\mu_B$ .

The precise statement

The Bohr magneton is defined as:

\mu_B = \frac{e\hbar}{2m_e}

where:

$e$ is the elementary charge ( $1.602 \times 10^{-19}$ C)
$\hbar$ is the reduced Planck constant ( $h/2\pi = 1.055 \times 10^{-34}$ J·s)
$m_e$ is the electron mass ( $9.109 \times 10^{-31}$ kg)

Plugging these numbers gives:

\mu_B = 9.274 \times 10^{-24} \text{ J/T}

Important

The unit J/T (joule per tesla) is the unit of magnetic moment. It tells you the energy of interaction when the magnet is placed in a magnetic field: $E = -\vec{\mu} \cdot \vec{B}$ .

Why this combination?

The formula $e\hbar/2m_e$ is not pulled from thin air. It comes directly from classical physics applied to a quantum orbit. For an electron in a circular orbit of radius $r$ with speed $v$ :

Orbital angular momentum: $L = m_e v r$
Current: $I = \frac{e}{T} = \frac{ev}{2\pi r}$ (charge per period)
Magnetic moment of a current loop: $\mu = I \times \text{area} = \frac{ev}{2\pi r} \cdot \pi r^2 = \frac{evr}{2}$

Now compare $\mu$ and $L$ :

\mu = \frac{evr}{2} = \frac{e}{2m_e} (m_e v r) = \frac{e}{2m_e} L

So the magnetic moment is $\frac{e}{2m_e}$ times the angular momentum. In quantum mechanics, angular momentum comes in units of $\hbar$ . So the smallest non-zero magnetic moment from orbital motion is when $L = \hbar$ , giving $\mu = e\hbar/2m_e$ .

Note

| Quantity | Symbol | Value |

|----------|--------|-------|

| Bohr magneton | $\mu_B$ | $9.274 \times 10^{-24}$ J/T |

| In eV/T | $\mu_B$ | $5.788 \times 10^{-5}$ eV/T |

| In cm $^{-1}$ /T | $\mu_B/hc$ | $0.467$ cm $^{-1}$ /T |

What it means in practice

When you study the Zeeman effect (splitting of spectral lines in a magnetic field), the energy shift of an atomic level is typically $\mu_B B$ times some small integer. When you calculate the magnetic moment of an atom, you always get an answer that is a multiple of $\mu_B$ . It is the natural yardstick.

The Bohr magneton is to magnetic moments what the elementary charge $e$ is to electric charges — the fundamental quantum.

The Bohr magneton is a constant-based CBSE Class 12 Physics NCERT topic, commonly searched as Bohr magneton value and formula class 12 or Bohr magneton derivation important questions. As the natural unit of atomic magnetic moment, it links Magnetism and Matter to the atomic-physics topics also tested in JEE Main and NEET physics.

Given $r=5.29\times10^{-11}\ \text{m}$ , $v=2.19\times10^6\ \text{m/s}$ , $e=1.6\times10^{-19}\ \text{C}$ .

Equivalent current:

I = \frac{ev}{2\pi r} = \frac{(1.6\times10^{-19})(2.19\times10^6)}{2\pi(5.29\times10^{-11})}

Numerator: $1.6\times10^{-19}\times2.19\times10^6 = 3.504\times10^{-13}$ .

Denominator: $2\pi\times5.29\times10^{-11} = 3.324\times10^{-10}$ .

I = \frac{3.504\times10^{-13}}{3.324\times10^{-10}} \approx 1.054\times10^{-3}\ \text{A}

Orbital magnetic moment:

\mu_l = I\times\pi r^2 = (1.054\times10^{-3})\times\pi\times(5.29\times10^{-11})^2

$\pi r^2 = \pi\times2.798\times10^{-21} \approx 8.79\times10^{-21}\ \text{m}^2$ .

\mu_l \approx (1.054\times10^{-3})\times(8.79\times10^{-21}) \approx 9.27\times10^{-24}\ \text{A}\cdot\text{m}^2

This matches the Bohr magneton exactly, as it must -- the first Bohr orbit is precisely the orbit for which the quantised orbital moment equals $\mu_B$ (Section 1.3).

✓Final answer

\mu_l \approx 9.27\times10^{-24}\ \text{A}\cdot\text{m}^2

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2019Set 55/2/13 marks
$r$
›Reveal solutionSolution
The magnetic moment of an orbiting electron is derived from its equivalent current loop, giving $\mu = evr/2$ . Using Bohr's quantization $mvr = \hbar$ , the ground-state magnetic moment of hydrogen becomes $\mu = e\hbar/2m$ , which is one Bohr magneton.

The key insight is that a moving charge creates a current, and a current loop has a magnetic moment. For an electron orbiting a nucleus, we can treat its circular path as a tiny current loop. The magnetic moment depends on the current (charge per unit time passing a point) and the area of the loop. Once we have the classical expression, Bohr's quantum condition fixes the angular momentum, which then determines the magnetic moment for the ground state.

Let's work through this step by step.

Current due to the orbiting electron The electron of charge $e$ (magnitude) completes one revolution in time $T = 2\pi r / v$ . The current $I$ is the charge passing a point per unit time:

$I = \frac{e}{T} = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r}.$
This is the effective current in the loop.

Magnetic moment of a current loop For a planar loop of area $A$ , the magnetic moment $\mu$ is $I \times A$ , directed perpendicular to the plane (right-hand rule). Here the area is $\pi r^2$ , so

$\mu = I \cdot A = \left(\frac{ev}{2\pi r}\right) \cdot (\pi r^2) = \frac{evr}{2}.$
That's the first result: $\mu = evr/2$ .

Tip

Notice the factor $1/2$ comes from the geometry — the current expression already has a $1/(2\pi r)$ , which cancels one factor of $r$ from the area, leaving $r/2$ . This is a clean, cancellation-heavy derivation.

Relating to angular momentum The orbital angular momentum of the electron (magnitude) is $L = mvr$ , where $m$ is the electron mass. So we can write

$\mu = \frac{e}{2m} (mvr) = \frac{e}{2m} L.$
This shows that the magnetic moment is proportional to the angular momentum, with the constant $e/(2m)$ called the gyromagnetic ratio.

Applying Bohr's quantization Bohr's postulate says that angular momentum in a stationary orbit is an integer multiple of $\hbar = h/(2\pi)$ :

$L = n\hbar, \quad n = 1, 2, 3, \dots$
For the ground state of hydrogen, $n = 1$ , so $L = \hbar$ .

Ground-state magnetic moment Substituting $L = \hbar$ into $\mu = (e/2m)L$ gives

$\mu = \frac{e}{2m} \cdot \hbar = \frac{e\hbar}{2m}.$
This quantity is known as the Bohr magneton, denoted $\mu_B$ . It is the fundamental unit of magnetic moment for atomic-scale currents.

Watch out

A common mistake is to forget that the electron's charge is negative. The magnitude of the magnetic moment is $evr/2$ , but the direction is opposite to the angular momentum vector because the electron has negative charge. In vector form: $\vec{\mu} = -\frac{e}{2m}\vec{L}$ . For the ground state, the magnitude is $\mu_B$ , but the vector points opposite to $\vec{L}$ .

$\mu_{\text{ground state}} = \frac{e\hbar}{2m} = \mu_B \approx 9.27 \times 10^{-24} \, \text{J/T}$
✓Final answer
$\boxed{\frac{e\hbar}{2m}}$

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