Q 12-11-173JEE MainJEE Main 2018 (16 Apr, Shift 1)Easy
The de-Broglie wavelength $(\lambda_B)$ associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state $(\lambda_G)$ by:
Answer: (A) $\lambda_B = 3\lambda_G$
Bohr's condition $2\pi r_n = n\lambda$, with $r_n \propto n^2$, gives $\lambda \propto n$.
The second excited state is $n = 3$ and the ground state is $n = 1$, so
$$\lambda_B = 3\lambda_G$$
Solution by Sreeraj P, M.Sc Physics