Q 12-12-118JEE MainJEE Main 2019 (12 Apr, Shift 2)Easy
Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65$ Å). The de Broglie wavelength of this electron is:
Answer: (C) $9.7$ Å
Second excited state: $n = 3$. Bohr's condition means $n$ wavelengths fit the orbit:
$$\lambda = \frac{2\pi r}{n} = \frac{2\pi\times4.65}{3} \approx 9.7\ \text{Å}$$
Solution by Sreeraj P, M.Sc Physics