Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65

Question

Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 \AA$ ). The de-Broglie wavelength of this electron is $\_\_\_\_$ Å.

Professor PC Thomas Classes · Answer

(9.7) : For second excited state, $n=3$ \ $ \begin{aligned} & 2 \pi r_n=n \lambda_n \ 	herefore \quad & \lambda_3=\frac{2 \pi r_3}{3}=\frac{2 \pi\left(4.65 	imes 10^{-10}ight)}{3}=9.7 \AA \end{aligned} $