(A) πa₀³ / 6
(B) 3πa₀
(C) 6πa₀
(D) πa₀³ / 3
Correct Answer: 6πa₀
According to Bohr–de Broglie condition:
2πrₙ = nλ
⇒ λ = 2πrₙ / n
Radius of nth orbit:
rₙ = n²a₀
For 3rd orbit (n = 3):
r₃ = 3²a₀ = 9a₀
λ = (2π × 9a₀) / 3
λ = 6πa₀
Hence, de Broglie wavelength of electron in 3rd orbit is 6πa₀.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.