Question 19 of 200
An electron in the ground state of the hydrogen atom has the orbital radius of while that for the electron in third excited state is . The ratio of the de Broglie wavelengths of electron in the excited state to that in the ground state is
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Answer: (D) 4
$\begin{aligned}
& \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \\
& \mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}} \\
& \lambda=\frac{2 \pi \mathrm{rh}}{\mathrm{nh}} \\
& \lambda \propto \frac{\mathrm{r}}{\mathrm{n}} \\
& \frac{\lambda_1}{\lambda_4}=\frac{\mathrm{r}_1 \mathrm{n}_4}{\mathrm{n}_1 \mathrm{r}_4}=\frac{5.3 \times 10^{-11} \times 4}{1 \times 84.8 \times 10^{-11}} \\
& \frac{\lambda_1}{\lambda_4}=\frac{1}{4}
\end{aligned}$