Magnetic moment due to the motion of the electron in nth energy state…

The correct answer is Option (1) → \( n \)

An orbiting electron is a tiny current loop, and its orbital magnetic moment is proportional to its angular momentum:

    \( \displaystyle \mu = \frac{e}{2m}\,L \)

By Bohr's quantisation condition, \( L = \displaystyle\frac{nh}{2\pi} \). So

    \( \displaystyle \mu = \frac{e}{2m}\times\frac{nh}{2\pi} = n\,\frac{eh}{4\pi m} \;\Rightarrow\; \mu \propto n \)

Shortcut: From \( \mu = evr/2 \) with \( v \propto 1/n \) and \( r \propto n^{2} \): \( \mu \propto (1/n)(n^{2}) = n \).

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
The Magnetic Dipole Moment of a Revolving Electron – “… There will be a magnetic moment, usually denoted by \( \mu_l \), associated with this circulating current. … Multiplying and dividing the right-hand side of the above expression by the electron mass m_e, we have,

    \( \displaystyle \mu_l = \frac{e}{2m_e}\left(m_e v r\right) = \frac{e}{2m_e}\, l \qquad [4.34(a)] \)

Here, l is the magnitude of the angular momentum of the electron about the central nucleus (“orbital” angular momentum).”

The Magnetic Dipole Moment of a Revolving Electron – “… Bohr hypothesised that the angular momentum assumes a discrete set of values, namely,

    \( \displaystyle l = \frac{nh}{2\pi} \qquad (4.36) \)

where n is a natural number, n = 1, 2, 3, ....”