Q 12-12-004NEETNEET 2025Top questionMedium
De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius $= 0.052$ nm)
Answer: (B) $0.67$ nm
Concept. Bohr's condition means a whole number of de Broglie wavelengths fits around the orbit: $2\pi r_n = n\lambda$.
Step 1. Radius of the $n = 2$ orbit. $r_2 = n^2a_0 = 4 \times 0.052 = 0.208$ nm.
Step 2. Wavelength.
$$\lambda = \frac{2\pi r_2}{2} = \pi \times 0.208 \approx 0.65\ \text{nm}$$
The closest option is $0.67$ nm.
Answer. $0.67$ nm (Option B).
Solution by Sreeraj P, M.Sc Physics