Physics Q18 – NEET-UG 2025

Home Physics Dual Nature of MatterQuestion #77

A model for quantized motion of an electron in a uniform magnetic field

B

states that the flux passing through the orbit of the electron is

n (h / e)

where

n

is an integer,

h

is Planck's constant and

e

is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (

m

is the mass of the electron)

Clue 1 of 3: Conceptual Understanding

The problem introduces a specific quantization condition for magnetic flux, $Φ = n (h / e)$ , which is crucial for determining the electron's orbital properties in the lowest energy state ( $n = 1$ ).

🧠 Full Solution Path

● Video Walkthrough

Step 1: Apply Flux Quantization and Determine Orbital Radius✦ Active

According to the model, the magnetic flux $Φ$ through the electron's orbit is quantized as $Φ = n \frac{h}{e}$ . For the lowest energy state, we take $n = 1$ , so $Φ = \frac{h}{e}$ .

The magnetic flux is also given by $Φ = B A$ , where $B$ is the uniform magnetic field and $A = π r^{2}$ is the area of the circular orbit. Equating these two expressions for flux:

B (π r^{2}) = \frac{h}{e}

From this, we can express the square of the orbital radius:

r^{2} = \frac{h}{π e B}

Step 2: Determine Angular Momentum○ Expand

For an electron orbiting in a magnetic field, the magnetic force provides the necessary centripetal force:

\frac{m v^{2}}{r} = e v B

This equation can be rearranged to relate $m v$ to $r$ :

m v = e B r

The angular momentum $L$ of the electron is defined as $L = m v r$ . Substituting $m v = e B r$ into the angular momentum definition:

L = (e B r) r = e B r^{2}

Now, substitute the expression for $r^{2}$ from Step 1 into the equation for $L$ :

L = e B (\frac{h}{π e B}) = \frac{h}{π}

💡 Teacher's Secret Hint

Remember that $m$ is the mass of the electron, $v$ is its speed, and $r$ is the orbital radius.

Step 3: Calculate Magnetic Moment○ Expand

The magnetic moment $μ$ of an orbiting electron is related to its angular momentum $L$ by the formula:

μ = \frac{e L}{2 m}

Substitute the value of $L = \frac{h}{π}$ found in Step 2 into this formula:

μ = \frac{e}{2 m} (\frac{h}{π}) = \frac{e h}{2 π m}

This result corresponds to option (2).

💡 Teacher's Secret Hint

This value is similar to the Bohr magneton, but derived from a specific flux quantization condition given in the problem.

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Physics Question 18 – NEET-UG 2025

Clue 1 of 3: Conceptual Understanding

🧠 Full Solution Path