JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2025
- Easy
- MCQ
An electron in the ground state of the hydrogen atom has the orbital radius of \(5.3 \times 10^{-11} \mathrm{~m}\) while that for the electron in third excited state is \(8.48 \times 10^{-10} \mathrm{~m}\). The ratio of the de Broglie wavelengths of electron in the excited state to that in the ground state is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\)
\(\begin{aligned}
& \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \\
& \mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}} \\
& \lambda=\frac{2 \pi \mathrm{rh}}{\mathrm{nh}} \\
& \lambda \propto \frac{\mathrm{r}}{\mathrm{n}} \\
& \frac{\lambda_1}{\lambda_4}=\frac{\mathrm{r}_1 \mathrm{n}_4}{\mathrm{n}_1 \mathrm{r}_4}=\frac{5.3 \times 10^{-11} \times 4}{1 \times 84.8 \times 10^{-11}} \\
& \frac{\lambda_1}{\lambda_4}=\frac{1}{4}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \\
& \mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}} \\
& \lambda=\frac{2 \pi \mathrm{rh}}{\mathrm{nh}} \\
& \lambda \propto \frac{\mathrm{r}}{\mathrm{n}} \\
& \frac{\lambda_1}{\lambda_4}=\frac{\mathrm{r}_1 \mathrm{n}_4}{\mathrm{n}_1 \mathrm{r}_4}=\frac{5.3 \times 10^{-11} \times 4}{1 \times 84.8 \times 10^{-11}} \\
& \frac{\lambda_1}{\lambda_4}=\frac{1}{4}
\end{aligned}\)
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