An electron in the ground state of the hydrogen atom has the orbital radius of 5.3 × 10^ -11 ~m while that for the electron in third excited state is 8.48 × 10^ -10 ~m . The ratio of the de Broglie wavelengths of

Correct answer: D. 4 λ= h mv aligned & mvr= nh 2 π & mv= nh 2 π r & λ= 2 π rh nh & λ r n & (λ_1)/(λ_4)= r_1 n_4 n_1 r_4 = 5.3 × 10^ -11 × 4 1 × 84.8 × 10^ -11 & (λ_1)/(λ_4)=(1)/(4) aligned

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2025 · 22 January · Morning Shift · Physics · Dual Nature of Matter

An electron in the ground state of the hydrogen atom has the...

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

Options

Correct answer

D. 4

Step-by-step solution

\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}

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