Q. An electron in the ground state of the hydrogen atom has the orbital radius of $5.3\times10^{-11}$ m while that for the electron in third excited state is $8.48\times10^{-10}$ m. The ratio of the de Broglie wavelengths of electron in the excited state to that in the ground state is

A) 3
B) 16
C) 9
D) 4
Explanation

[Solution by Dr. Abubacker] $\lambda=\frac{2\pi r}{n}$. The third excited state has $n=4$ and $r_4=16r_1$, so $\frac{\lambda_4}{\lambda_1}=\frac{16/4}{1}=4$. Answer (4).

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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