JEE Main 2025 · 29 January · morning · Single correct · +4 / −1
If a₀ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength (λ) of the electron present in the second orbit of hydrogen atom? [n : any integer]
- (a) (8 π a₀)/n
- (b) (2 a₀)/(n π)
- (c) (n a₀)/(4 π)
- (d) (4 π a₀)/n
Answer and solution
Answer (a) (8 π a₀)/n
Solution
Chemistry ma’am · 4 steps
Find the second orbit's radius first, then fit whole wavelengths around its circumference.
- Radius of orbit 2. In hydrogen, rₙ = n² a₀. For the second orbit this gives r₂ = 2² a₀ = 4a₀.
- Standing wave condition. In Bohr's model the electron wave must close on itself, so a whole number of wavelengths fits exactly around the orbit. This is Bohr's quantisation rule mvr = nh/2π combined with λ = h/mv.
nλ = 2π r₂ = 2π(4a₀) = 8π a₀
- Solve for λ. Divide by n to get λ = (8π a₀)/n. For the second orbit n = 2, so this equals 4π a₀. The options keep n as a symbol, so the matching form is (8π a₀)/n.
- So. The de Broglie wavelength is (8π a₀)/n, which is option (a).
The idea to keep. In a Bohr orbit, a whole number of de Broglie wavelengths fits around the circumference: nλ = 2π r. Remember that the radius grows as n², not as n.
Answer from the NTA final answer key. Solution by PrepWiser.