Q 12-12-033JEE MainJEE Main 2025 (22 Jan, Shift 1)Easy
An electron in the ground state of the hydrogen atom has the orbital radius of $5.3\times10^{-11}\ \text{m}$ while that for the electron in third excited state is $8.48\times10^{-10}\ \text{m}$. The ratio of the de Broglie wavelengths of electron in the excited state to that in the ground state is
Answer: (D) $4$
Bohr's condition $2\pi r_n = n\lambda_n$ gives $\lambda_n = \dfrac{2\pi r_n}{n}$.
The third excited state is $n = 4$ (check: $r_4/r_1 = 8.48\times10^{-10}/5.3\times10^{-11} = 16 = 4^2$).
$$\frac{\lambda_4}{\lambda_1} = \frac{r_4/4}{r_1/1} = \frac{16}{4} = 4$$
Solution by Sreeraj P, M.Sc Physics