In a hydrogen atom, an electron is moving in the nth permitted orbit…

The correct answer is Option (3) → \( \displaystyle\frac{2\pi r_n}{n} \)

The wavelength of the electron is the de Broglie wavelength \( \lambda = h/mv \), and in the nth Bohr orbit the angular momentum is quantised, \( mvr_n = nh/2\pi \) (Bohr’s second postulate).

    \( \displaystyle mv = \frac{nh}{2\pi r_n} \)

    \( \displaystyle \lambda = \frac{h}{mv} = \frac{h}{nh/2\pi r_n} = \frac{2\pi r_n}{n} \)

This is the same as saying that the circumference of the orbit holds a whole number of wavelengths, \( 2\pi r_n = n\lambda \).

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27):
De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation – “… For an electron moving in nth circular orbit of radius r_n, the total distance is the circumference of the orbit, 2πr_n. Thus

    \( \displaystyle 2\pi r_n = n\lambda, \quad n = 1, 2, 3... \qquad (12.12) \)

Figure 12.8 illustrates a standing particle wave on a circular orbit for n = 4, i.e., 2πr_n = 4λ, where λ is the de Broglie wavelength of the electron moving in nth orbit.”