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
Radius of the $n = 2$ orbit: $r_2 = n^2a_0 = 4 \times 0.052 = 0.208$ nm.
Bohr's condition means the orbit holds a whole number of wavelengths: $2\pi r_n = n\lambda$.
$$\lambda = \frac{2\pi r_2}{2} = \pi \times 0.208 \approx 0.65\ \text{nm}$$
The closest option is $0.67$ nm.
Solution by Sreeraj P, M.Sc Physics